Prove Sum Approximation Theorem

[Fredrik]

$$\sum_{k=1}^{\infty} b_{k}=S; \space \space \space \mbox{Let } m \geq M \space \space \& \space \space \varepsilon > 0\\
\Longrightarrow |\sum_{k=1}^{m}-S|<\varepsilon$$

It's going to be useful to have a symbol for the sum $\sum_{k=1}^\infty b_k$, so I like that you assigned the symbol S right away.

When you say "let $m\geq M$", this is interpreted as "let m be an arbitrary integer such that $m\geq M$". This isn't a meaningful thing to say unless you have already specified what M is. If you want both m and M to be arbitrary apart from the fact that they satisfy that inequality, you have to say something like "let m and M be integers such that $m\geq M$". (This would be a bad idea in this proof).

It's correct to start with "let $\varepsilon>0$", but you can't leave M arbitrary. You need to let M be a positive integer such that the implication
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}$$
holds for all positive integers m such that $m\geq M$. The existence of such an M is implied by the definition of the statement $\sum_{k=1}^\infty b_k=S$. The reason why we're saying that M is such an integer is that we're preparing to use the assumption that $\sum_{k=1}^\infty b_k=S$. Why the |c| in the denominator? The reason will become clear when we get to the end of the proof. For now, just note that the definition of limit allows us to put that |c| there.

I'm pretty sure it's also valid to write down
$$c \in \mathbb{R} \space \space |\sum_{k=1}^{m}cb_{k}-cS| < \varepsilon $$

This looks a lot like the statement you want to prove, and it's never OK to just write that down as if it's known to be a true statement. Also, since you have left M unspecified, the condition $m\geq M$ may not be sufficient to ensure that this inequality holds.

What you need to do is to let $m$ be an (arbitrary) integer such that $m\geq M$ (in more difficult proofs, you would have to replace M with a different number at this point*), write down only the left-hand side of the inequality you want to prove, i.e. $\big|\sum_{k=1}^m cb_k- cS\big|$, and then work with that expression (using what we said at the start of the proof), until you have found the inequality we want, i.e.
$$\left|\sum_{k=1}^m cb_k- cS\right|<\varepsilon.$$

*) For an example, see the proof in post #24, where I used $\max\{M,N+1\}$ instead of $M$.

 

[Potatochip911]

This looks a lot like the statement you want to prove, and it's never OK to just write that down as if it's known to be a true statement. Also, since you have left M unspecified, the condition $m\geq M$ may not be sufficient to ensure that this inequality holds.

Okay I am slightly confused now since I was able to get $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$ by starting with $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$, if I can't assume this is true I'm not sure how to end up with $|c|$ in the denominator.

 

[Fredrik]

Okay I am slightly confused now since I was able to get $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$ by starting with $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$, if I can't assume this is true I'm not sure how to end up with $|c|$ in the denominator.

You're not supposed to end up with $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$. You simply define M as a positive integer such that this inequality holds for all $m\geq M$. Then you use that inequality to prove that $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$ for all $m$ such that $m\geq M$.

Recall that the statement $\sum_{k=1}^\infty b_k=S$ means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m. $$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\varepsilon.$$ The "for all" is the key. If you let $\varepsilon$ be an arbitrary positive real number, then $\varepsilon/|c|$ is a positive real number too. So the statement $\sum_{k=1}^\infty b_k=S$ implies that there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$

 

[Potatochip911]

You're not supposed to end up with $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$. You simply define M as a positive integer such that this inequality holds for all $m\geq M$. Then you use that inequality to prove that $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$ for all $m$ such that $m\geq M$.

Recall that the statement $\sum_{k=1}^\infty b_k=S$ means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m. $$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\varepsilon.$$ The "for all" is the key. If you let $\varepsilon$ be an arbitrary positive real number, then $\varepsilon/|c|$ is a positive real number too. So the statement $\sum_{k=1}^\infty b_k=S$ implies that there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$

So to prove this I want to assume that for all $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$ with $M\in\mathbb{N}, M\neq0$ and now it's also valid to assume this holds? $$m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|} \space \space \mbox{which leads to} \\
|c||\sum_{k=1}^{m}b_{k}-S|<\varepsilon \\
|c\sum_{k=1}^{m}b_{k}-cS|<\varepsilon \\
|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon $$

 

[Fredrik]

So to prove this I want to assume that for all $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$

It's a bit odd to use the "for all" and the implication arrow like that. It would be better to say e.g. "for all $m\geq M$, we have $|\sum_{k=1}^{m}b_{k}-S|<\varepsilon$" or "for all $m$, we have $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$".

We don't assume that this implication holds. Our assumption is that $\sum_{k=1}^\infty b_k$ is convergent. If we denote the sum of the series by $S$, then this assumption implies that there's an M such that the implication holds for all m.

and now it's also valid to assume this holds? $$m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|} \space \space \mbox{which leads to} \\
|c||\sum_{k=1}^{m}b_{k}-S|<\varepsilon \\
|c\sum_{k=1}^{m}b_{k}-cS|<\varepsilon \\
|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon $$

The first line is the assumption. The last line is a consequence of that assumption.

OK, let's clean this up a little. Here's how I would type the proof:

Let $\varepsilon>0$. Let M be a positive integer such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$ For all $m\geq M$, we have
$$\left|\sum_{k=1}^m cb_k-cS\right| = |c|\left|\sum_{k=1}^m b_k -S\right|<|c|\frac{\varepsilon}{|c|}=\varepsilon.$$

I think the best way to approach proofs like this is to start by looking at the statement you want to prove. (This is when you're trying to find the proof). That statement is $\sum_{k=1}^\infty cb_k=cS$. It means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m cb_k-cS\right| <\varepsilon.$$ Because of the "for all" at the start of that statement, the proof should start with "let $\varepsilon>0$". Now the question is how to choose M. Since the inequality that's part of the the statement that we want to prove is equivalent to $\big|\sum_{k=1}^m b_k-S\big|<\frac{\varepsilon}{|c|}$, all we have to do is to choose M such that this inequality holds for all m such that $m\geq M$. When you type up the proof, you start with this choice of M, and then use it to prove the statement we want to prove.

 

[Potatochip911]

Now trying to prove the first theorem:

Theorem 1: If $(S_{n})_{n=1}^{\infty}$is convergent and c is a real number, then $(S_{k}-c)_{n=1}^{\infty}$ is convergent, and we have $$\lim_{n} (S_{n}-c)=\lim_{n} S_{n}-c$$

let $S=\sum_{n=1}^{\infty}S_{n}$ and $\varepsilon>0$
$$\mbox{For all } m\geq M \Rightarrow |\sum_{n=1}^{m}S_{n}-S|<\varepsilon$$
I'm not sure where to go next and this definition isn't making a lot of sense to me right now. To check if a sequence is converging we can use the preliminary test to see if $\lim_{n\to\infty}a_{n}\rightarrow 0$ I don't see how this is possible when we have a real number c in each term. Also $\sum_{n=1}^{\infty}(S_{n}-c)=\sum_{n=1}^{\infty}S_{n}-\sum_{n=1}^{\infty}c$ and since $c$ is a constant $$\sum_{n=1}^{\infty}c=\lim_{n\to\infty} nc$$ which is clearly going to infinity. I feel like I'm probably misunderstanding how this sum works and that $\sum_{n=1}^{\infty}(S_{n}-c)=(\sum_{n=1}^{\infty}S_{n})-c$

 

[Fredrik]

Now trying to prove the first theorem:

let $S=\sum_{n=1}^{\infty}S_{n}$ and $\varepsilon>0$

$(S_n)_{n=1}^\infty$ is an arbitrary convergent sequence in this theorem, so there are no sums involved. You should change this to "let $S=\lim_n S_n$ and $\varepsilon>0$".

Now to find the rest of the proof, think about what exactly the statement that you want to prove means. See post #18 for the definition of limit.

 

[Ray Vickson]

Okay I am slightly confused now since I was able to get $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$ by starting with $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$, if I can't assume this is true I'm not sure how to end up with $|c|$ in the denominator.

If you start with $|S_n - S| < \epsilon'$ you have $|cS_n - cS| < |c|\epsilon'$. Give $|c| \epsilon'$ the new name $\epsilon$, so you have $|cS_n - cS| < \epsilon$.

Now work backwards: you want to specify $\epsilon$, so take $\epsilon' = \epsilon/|c|$. You know there exists an $N = N(\epsilon')$ such that $|S_n - S| < \epsilon'$ for all $n \geq N$, so that means that $|cS_n - cS| < \epsilon$ for all $n \geq N$.

The real point is that whether you call the thing $\epsilon$ or $\eta$ or $\omega$ or $pqr$ is irrelevant; what matters is that if you want to get within a specified distance from $cS$ you can do it by ensuring that $n$ is large enough.

 

[Ray Vickson]

Now trying to prove the first theorem:

let $S=\sum_{n=1}^{\infty}S_{n}$ and $\varepsilon>0$
$$\mbox{For all } m\geq M \Rightarrow |\sum_{n=1}^{m}S_{n}-S|<\varepsilon$$
I'm not sure where to go next and this definition isn't making a lot of sense to me right now. To check if a sequence is converging we can use the preliminary test to see if $\lim_{n\to\infty}a_{n}\rightarrow 0$ I don't see how this is possible when we have a real number c in each term. Also $\sum_{n=1}^{\infty}(S_{n}-c)=\sum_{n=1}^{\infty}S_{n}-\sum_{n=1}^{\infty}c$ and since $c$ is a constant $$\sum_{n=1}^{\infty}c=\lim_{n\to\infty} nc$$ which is clearly going to infinity. I feel like I'm probably misunderstanding how this sum works and that $\sum_{n=1}^{\infty}(S_{n}-c)=(\sum_{n=1}^{\infty}S_{n})-c$

Yes, you have misunderstood: since $S_n = \sum_{k=1}^n a_k x^k$ we have $S_n - c = \left(\sum_{k=1}^n a_n x^n\right) - c$, NOT $\sum_{k=1}^n (a_k x^k - c)$, which is a totally different animal.

 

[Noctisdark]

|x^N| = |x|^N, now consider the sum over i from N+1 to a some M, then Σa*|x|^i = |x|^N+1 * a * (|x|^(M-N) - 1)/(1-|x|), if M goes to infinity then he result iwill havr and |x|^(-N), because |x|<1then |x|^(-N) and N > 0 then|x|^(-N) < |x|^(N+1), and taadaa

 

[Potatochip911]

$(S_n)_{n=1}^\infty$ is an arbitrary convergent sequence in this theorem, so there are no sums involved. You should change this to "let $S=\lim_n S_n$ and $\varepsilon>0$".

Now to find the rest of the proof, think about what exactly the statement that you want to prove means. See post #18 for the definition of limit.

So I should do something along the lines of this?
Let $S=\lim_{n} S_{n}$ and $\varepsilon>0$ Let c be a real number. Let n & N both be positive integers such that
$$n\geq N \Rightarrow |S_{n}|<\varepsilon+c \\
|S_{n}|-c<\varepsilon \\
|S_{n}-c|<\varepsilon \\
|S_{n}-c|\leq |S_{N}-c|<\varepsilon $$ I'm not too sure about this...

 

[Fredrik]

You don't have to say "let c be a real number", since that's stated in the theorem. Actually, it's confusing to say it at the start of the proof, since that suggests that now you're talking about a different c.

Your definition of N isn't useful. Suppose e.g. that the inequality holds when n=3 and when n>100, but for no other values of n. Then your statement allows n and N to both be 3.

Also, it's the wrong inequality. Can you take another look at the definition of limit and just tell me what the following statements mean:
$$\lim_n S_n=S,\qquad \lim (S_n-c)=T.$$

 

[Potatochip911]

You don't have to say "let c be a real number", since that's stated in the theorem. Actually, it's confusing to say it at the start of the proof, since that suggests that now you're talking about a different c.

Your definition of N isn't useful. Suppose e.g. that the inequality holds when n=3 and when n>100, but for no other values of n. Then your statement allows n and N to both be 3.

Also, it's the wrong inequality. Can you take another look at the definition of limit and just tell me what the following statements mean:
$$\lim_n S_n=S,\qquad \lim (S_n-c)=T.$$

For $\lim_{n} S_{n}=S$ S is the limit of $S_{n}$ and for $\lim(S_{n}-c)=T$ T is the limit of $S_{n}-c$, sorry if I wasn't supposed to write down what they mean literally but I wasn't sure what else to do. I'm kinda confused by what $S_{n}$ represents, is it something like in post #18?

 

[Fredrik]

For $\lim_{n} S_{n}=S$ S is the limit of $S_{n}$ and for $\lim(S_{n}-c)=T$ T is the limit of $S_{n}-c$, sorry if I wasn't supposed to write down what they mean literally but I wasn't sure what else to do. I'm kinda confused by what $S_{n}$ represents, is it something like in post #18?

$S_n$ denotes the $n$th term of the sequence $(S_n)_{n=1}^\infty$. What I'm asking you to do is to apply the definition of limit (without making any sort of assumptions about what sort of sequence we're dealing with...in particular you shouldn't assume that it's a sequence of partial sums of a series). That definition is what tells you what these statements mean. It's impossible to do any of these proofs without understanding how to apply the definition.

 

[Potatochip911]

$S_n$ denotes the $n$th term of the sequence $(S_n)_{n=1}^\infty$. What I'm asking you to do is to apply the definition of limit (without making any sort of assumptions about what sort of sequence we're dealing with...in particular you shouldn't assume that it's a sequence of partial sums of a series). That definition is what tells you what these statements mean. It's impossible to do any of these proofs without understanding how to apply the definition.

I'm pretty sure the definition of the limit is $|f(x)-L|<\varepsilon$ in this particular case $f(x)$ would be $S_{n}$ and $L$ would be $\lim_{n} S_{n}$ Then $$|S_{n}-\lim_{n}S_{n}|<\varepsilon$$

 

[Fredrik]

That statement that you're supposed to explain is $\lim_n S_n=S$. Obviously, it means that S is a limit of the sequence $(S_n)_{n=1}^\infty$. The difficult part is to explain what that means. The explanation should have the symbol S where you wrote $\lim_n S_n$, because you want to say that S is a limit of the sequence, not that the limit of the sequence is a limit of the sequence.

You have the right inequality this time, but you haven't fully explained what the statement means unless you say what $\varepsilon$ is, or mentioned the special positive integer N.

This is the exact meaning of the claim that S is a limit of the sequence:

For all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |S_n-S|<\varepsilon.$$
Now can you explain what $\lim_n (S_n-c)=T$ means?

 

[Potatochip911]

*Editing*
You have the right inequality this time, but you haven't fully explained what the statement means unless you say what $\varepsilon$ is, or mentioned the special positive integer N.

The statement $\lim_n S_n=S$ says that S is a limit of the sequence $(S_n)_{n=1}^\infty$, and that means exactly this:

For all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |S_n-S|<\varepsilon.$$
Now can you explain what $\lim_n (S_n-c)=T$ means?

I think the inequality is: Let $\varepsilon>0$ and n is a positive integer such that $$n\geq N \Rightarrow|\lim_{n}(S_{n}-c)-T|<\varepsilon \\
|\lim_{n}S_{n}-\lim_{n}c-T|<\varepsilon \\
|\lim_{n}S_{n}-c-T|<\varepsilon
$$

 

[Fredrik]

You still need to say that N is a positive integer such that the implication holds for all n.

The inequality is not correct. $\lim_n(S_n-c)$ is equal to $T$, so it's pointless to say that the distance between $\lim_n(S_n-c)$ and $T$ is less than $\varepsilon$. Look at the definition of limit again, or my explanation of what $\lim_n S_n=S$ means in my previous post. There's no $\lim$ in the inequality. My statement specifies what the terms of the sequence must satisfy for the sequence $(S_n)_{n=1}^\infty$ to have the limit S.

What you need to say is almost exactly the same thing I said, but with changes that reflect that we're dealing with the sequence $(S_n-c)_{n=1}^\infty$ instead of $(S_n)_{n=1}^\infty$, and that the limit of this sequence is denoted by T instead of S.

Edit: Since you're still struggling with the definition, I suggest that you first make sure that you understand the "wordy" version of it:

Let S be a sequence. A real number x is said to be a limit of S, if every open interval with x at the center contains all but a finite number of terms of S.​

Then make sure that you understand that the version of the definition that involves the $\varepsilon$ symbol is just saying exactly that, only in even more precise language.

 

[Fredrik]

Hey, you didn't give up, did you? Are you ready to prove that $\lim_n S_n=S$ implies that $\lim_n (S_n-c)=S-c$?

 

[Potatochip911]

Hey, you didn't give up, did you? Are you ready to prove that $\lim_n S_n=S$ implies that $\lim_n (S_n-c)=S-c$?

My summer classes started on Monday so I have been kind of busy with that and despite rereading post #48 multiple times over the last few days I couldn't figure out what to do. I am having trouble grasping what $(S_{n})_{n=1}^{\infty}$ is, is it the same as $\sum_{n=1}^{\infty}S_{n}$?

 

[Fredrik]

My summer classes started on Monday so I have been kind of busy with that and despite rereading post #48 multiple times over the last few days I couldn't figure out what to do. I am having trouble grasping what $(S_{n})_{n=1}^{\infty}$ is, is it the same as $\sum_{n=1}^{\infty}S_{n}$?

No, it's just an arbitrary infinite sequence $S_1,S_2,\dots$. Post #18 is the one you need to understand. The proof of the claim that $\lim_nS_n=S$ implies $\lim(S_n-c)=S-c$ is a very straightforward application of that definition. Don't give up. This concept is the very core of calculus and analysis. As long as you don't understand it, you will always find analysis unreasonably hard.

The definition of "limit" is just this:

Let σ be a sequence of real numbers. A real number x is said to be a limit of σ, if for all positive real numbers ε, all but a finite number of the terms of σ are at a distance from x that's less than ε.​

So to prove that $\lim_n (S_n-c)=S-c$ (given that $\lim_n S_n=S$) you start by letting ε be an arbitrary real number, and now you need to prove that that all but a finite number of the $S_n-c$ are at a distance from $S_n-c$ that's less than ε. And you are allowed to use that all but a finite number of the $S_n$ are at a distance from $S$ that's less than ε.

 

[Potatochip911]

No, it's just an arbitrary infinite sequence $S_1,S_2,\dots$. Post #18 is the one you need to understand. The proof of the claim that $\lim_nS_n=S$ implies $\lim(S_n-c)=S-c$ is a very straightforward application of that definition. Don't give up. This concept is the very core of calculus and analysis. As long as you don't understand it, you will always find analysis unreasonably hard.

The definition of "limit" is just this:

Let σ be a sequence of real numbers. A real number x is said to be a limit of σ, if for all positive real numbers ε, all but a finite number of the terms of σ are at a distance from x that's less than ε.​

So to prove that $\lim_n (S_n-c)=S-c$ (given that $\lim_n S_n=S$) you start by letting ε be an arbitrary real number, and now you need to prove that that all but a finite number of the $S_n-c$ are at a distance from $S_n-c$ that's less than ε. And you are allowed to use that all but a finite number of the $S_n$ are at a distance from $S$ that's less than ε.

Let $\varepsilon>0$ and Let $\varepsilon'>0$
$$|S_{n}-\lim_{n}S_{n}|<\varepsilon' \\
|S_{n}-\lim_{n}S_{n}|-2c<\varepsilon'-2c \\
\mbox{for } |S_{n}-\lim_{n}S_{n}| > 0 \Rightarrow |S_{n}-c-\lim_{n}S_{n}-c|<\varepsilon'-2c \\
\varepsilon=\varepsilon'-2c \\
|S_{n}-c-\lim_{n}S_{n}-c|<\varepsilon \\
|S_{n}-c-\lim_{n}(S_{n}-c)|<\varepsilon
$$

 

[Fredrik]

There are several mistakes in there. In particular, you can't use two different epsilons, and there are parentheses missing. I will do this one too. Maybe the explanation will give you what you need to be able to solve another problem on your own.

The statement we want to prove is $\lim_n (S_n-c)=S-c$. Before we prove the statement, we must understand what it says. It says that for all $\varepsilon>0$, all but a finite number of terms of the sequence $(S_n-c)_{n=1}^\infty$ are at a distance from S-c that's less than $\varepsilon$. Let's rephrase this again, this time spelling out exactly what "all but a finite" means: For all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |(S_n-c)-(S-c)|<\varepsilon.$$ This is the statement we want to prove. Since it's a "for all ε>0" statement, the proof should start with "Let ε>0". Now we need to find a positive integer N such that the implication above holds for all positive integers n. First note that the left-hand side of the inequality is equal to $|S_n-S|$. So we're looking for a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |S_n-S|<\varepsilon.$$ Does such an N exist? Yes! That's exactly what the assumption $\lim_n S_n=S$ is saying. Now we just need to type it up:

Theorem: If $\lim_n S_n=S$, then $\lim_n(S_n-c)=S-c$.

Proof: Let $\varepsilon>0$. Let N be a positive integer such that the following implication holds for all positive integers n
$$n\geq N\ \Rightarrow\ |S_n-S|<\varepsilon.$$ For all positive integers n such that $n\geq N$, we have
$$|(S_n-c)-(S-c)|=|S_n-S|<\varepsilon.$$
That's it. That's the entire proof. The first sentence uses the assumption $\lim_n S_n=S$ to choose N sufficiently large, and the second sentence proves that $\lim_n(S_n-c)=S-c$.

 

[Fredrik]

Did you understand this proof? I would still like to see you do at least one of these proofs on your own before I show you my solution of the problem you asked about. It can literally be any proof that involves the definition of limit. The following are two theorems that can be proved using the definition of limit. Can you prove any of them?

1. If x and y are both limits of $(x_n)_{n=1}^\infty$, then x=y.
2. Let c be a real number. If $(x_n)_{n=1}^\infty$ is convergent, then so is $(cx_n)_{n=1}^\infty$, and $\lim_n (cx_n)=c\lim x_n$.

(The phrase "is convergent" means "has a limit"). For the proof of the second theorem, I suggest that you denote $\lim_n x_n$ by $x$.

 

[Potatochip911]

Did you understand this proof? I would still like to see you do at least one of these proofs on your own before I show you my solution of the problem you asked about. It can literally be any proof that involves the definition of limit. The following are two theorems that can be proved using the definition of limit. Can you prove any of them?

1. If x and y are both limits of $(x_n)_{n=1}^\infty$, then x=y.
2. Let c be a real number. If $(x_n)_{n=1}^\infty$ is convergent, then so is $(cx_n)_{n=1}^\infty$, and $\lim_n (cx_n)=c\lim x_n$.

(The phrase "is convergent" means "has a limit"). For the proof of the second theorem, I suggest that you denote $\lim_n x_n$ by $x$.

Yea I understood your proof, quite disappointing I failed to see that since it was just a form of zero again. I need to think more about the first one but isn't the second one the same as what was proven in post 34/35?

 

[Fredrik]

Yea I understood your proof, quite disappointing I failed to see that since it was just a form of zero again. I need to think more about the first one but isn't the second one the same as what was proven in post 34/35?

Ah, yes, it's essentially the same. The only difference is that now it's an arbitrary sequence instead of a sequence of partial sums. The proof will be very similar.

 

[Potatochip911]

Ah, yes, it's essentially the same. The only difference is that now it's an arbitrary sequence instead of a sequence of partial sums. The proof will be very similar.

Okay for the second proof I have:
$x=\lim_{n}x_{n}$
Let $\varepsilon>0$ and let N and n be positive integers such that $$
n\geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|c||x_{n}-x|<\varepsilon \\
|cx_{n}-cx|<\varepsilon \\
|cx_{n}-c\lim_{n}x_{n}|<\varepsilon \\
|cx_{n}-\lim_{n}c\cdot\lim_{n}x_{n}|<\varepsilon \\
|cx_{n}-\lim_{n}(cx_{n})|<\varepsilon $$
I'm still working on the first one.

 

[Fredrik]

A few comments:

1. You might want to make it more clear that the equality $x=\lim_n x_n$ is the definition of x. It's sufficent to add the word "define" before the equality: "Define $x=\lim_n x_n$."

2. Your choice of N fails because you didn't make a "for all n" statement. This is my version of what needs to be said: Let N be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N \Rightarrow\ |x_n-x|<\frac{\varepsilon}{|c|}.$$
Suppose e.g. that for some specific choice of the sequence $(x_n)_{n=1}^\infty$, the smallest N with that property is the number 100. Then we have
\begin{align*}
&|x_{100}-x|<\frac{\varepsilon}{|c|}\\
&|x_{101}-x|<\frac{\varepsilon}{|c|}\\
&\vdots
\end{align*} but the same sequence may also be such that
$$|x_{3}-x|<\frac{\varepsilon}{|c|}.$$ For such a sequence, the statement in your proof attempt allows the possibility N=n=3.

3. When you write one statement on each line, the reader will interpret this as you saying that each statement (except the first) is true because the statement on the previous line is true. You wrote statements of the form

If A then B
C
D
​

where B implied C, and C implied D. This is interpreted as saying that "if A then B" implies C, not as saying that B implies C.

4. After the line that contains the implication arrow, it's not clear what you're doing. Yes, you're looking at inequalities that you can derive from the inequality in that first line, but why? What you need to think about here is this: What does the statement $\lim(cx_n)=cx$ mean? (What does the definition of limit say that it means?). The statement that the definition says is equivalent to $\lim(cx_n)=cx$ is the statement that you need to prove. But I don't see any indication that this is the statement that you're trying to prove.

5. In the fourth line, you're substituting $\lim_n x_n$ for x. If we would need to do this at any point in the proof, it would be kind of pointless to introduce the notation x in the first place.

6. In the fourth line, you're using that the limit of the constant sequence c,c,c,... is c. This is correct, but it's a result that we haven't proved here. There is no need to use any other theorems in this proof.

7. In the fifth line, you're using a theorem about products of limits that we also haven't proved. It would be especially bad to use this one without proof, because the theorem I asked you to prove is a special case of it.Of all these comments, the most important one by far is the part of number 4 that reminds you that in order to prove the claim $\lim_n(cx_n)=cx$, you have to know what it means. You should write it out for yourself on paper. Then you will have the statement that you need to prove right in front of you. I think that when you start doing this right, the other pieces will fall into place. In particular, you will see why the choice of N that you made is bad. That choice was supposed to enable you to complete the proof of that statement.

 

[Potatochip911]

A few comments:

1. You might want to make it more clear that the equality $x=\lim_n x_n$ is the definition of x. It's sufficent to add the word "define" before the equality: "Define $x=\lim_n x_n$."

2. Your choice of N fails because you didn't make a "for all n" statement. This is my version of what needs to be said: Let N be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N \Rightarrow\ |x_n-x|<\frac{\varepsilon}{|c|}.$$
Suppose e.g. that for some specific choice of the sequence $(x_n)_{n=1}^\infty$, the smallest N with that property is the number 100. Then we have
\begin{align*}
&|x_{100}-x|<\frac{\varepsilon}{|c|}\\
&|x_{101}-x|<\frac{\varepsilon}{|c|}\\
&\vdots
\end{align*} but the same sequence may also be such that
$$|x_{3}-x|<\frac{\varepsilon}{|c|}.$$ For such a sequence, the statement in your proof attempt allows the possibility N=n=3.

3. When you write one statement on each line, the reader will interpret this as you saying that each statement (except the first) is true because the statement on the previous line is true. You wrote statements of the form

If A then B
C
D
​

where B implied C, and C implied D. This is interpreted as saying that "if A then B" implies C, not as saying that B implies C.

4. After the line that contains the implication arrow, it's not clear what you're doing. Yes, you're looking at inequalities that you can derive from the inequality in that first line, but why? What you need to think about here is this: What does the statement $\lim(cx_n)=cx$ mean? (What does the definition of limit say that it means?). The statement that the definition says is equivalent to $\lim(cx_n)=cx$ is the statement that you need to prove. But I don't see any indication that this is the statement that you're trying to prove.

5. In the fourth line, you're substituting $\lim_n x_n$ for x. If we would need to do this at any point in the proof, it would be kind of pointless to introduce the notation x in the first place.

6. In the fourth line, you're using that the limit of the constant sequence c,c,c,... is c. This is correct, but it's a result that we haven't proved here. There is no need to use any other theorems in this proof.

7. In the fifth line, you're using a theorem about products of limits that we also haven't proved. It would be especially bad to use this one without proof, because the theorem I asked you to prove is a special case of it.Of all these comments, the most important one by far is the part of number 4 that reminds you that in order to prove the claim $\lim_n(cx_n)=cx$, you have to know what it means. You should write it out for yourself on paper. Then you will have the statement that you need to prove right in front of you. I think that when you start doing this right, the other pieces will fall into place. In particular, you will see why the choice of N that you made is bad. That choice was supposed to enable you to complete the proof of that statement.

The statement $\lim(cx_{n})=cx$ would mean $$|cx-\lim(cx_{n})|<\varepsilon$$ although I'm probably not allowed to just write this statement down.
Edit: Am I also supposed to just use the assumption $\lim(cx_{n})=c\lim(x_{n})$ so instead
$$
|cx-\lim(cx_{n})|=|cx-c\lim(x_{n})|=|c||x-\lim(x_{n})|<\varepsilon
$$

 

[Fredrik]

The statement $\lim(cx_{n})=cx$ would mean $$|cx-\lim(cx_{n})|<\varepsilon$$

This is incorrect. According to the definition of "limit", the statement means that for all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |cx_n-cx|<\varepsilon.$$
The first "for all", the "there exists" and the second "for all" are all essential parts of the statement. If you miss anyone of them, the proof will fail.

Also note that it wouldn't make much sense for the definition of limit to use the notation "lim" in the explanation of what a limit is. The definition specifies what it means to say that a number is a limit of a sequence. Once we have done that, and proved that no sequence has more than one limit, we can define the notation "lim" by saying that $\lim_n x_n$ denotes the unique limit of the sequence $(x_n)_{n=1}^\infty$.

Edit: Am I also supposed to just use the assumption $\lim(cx_{n})=c\lim(x_{n})$ so instead
$$
|cx-\lim(cx_{n})|=|cx-c\lim(x_{n})|=|c||x-\lim(x_{n})|<\varepsilon
$$

That's not an assumption. It's what you're trying to prove. So you absolutely cannnot use it here.

If you need to use it in some other proof, it wouldn't look like that. You would have to make it clear how you're using the "for all $\varepsilon>0$" (by saying "let $\varepsilon>0$"), how you're using the "there exists" (to make a choice of N that will be useful later) and how you're using the "for all n" (by making a "for all n" statement). It would be very similar to what I did in the first sentence of the proof in post #53. The statement would involve $|cx-cx_n|$, not $|cx-c\lim_n x_n|$. It wouldn't involve the "lim" notation at all, since it's a statement made by a definition that logically had to be stated before that notation was introduced.