Limit proof that 1=0 - Where is the error?

[krackers]

Limit proof that 1=0 -- Where is the error?

\lim_{n\to\infty } 1=\lim_{n\to\infty }\frac{n}{n}=\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n}=\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots=0

I have a feeling it is in the step where you split the limit of a sum into the sum of the limit of the terms, but I do not have any exact explanation as to why. Can anyone help?

 

[D H]

The limit of a sum is the sum of the limits on the individual terms if the number of terms is finite and if each of the terms has a limit. This is the algebraic limit theorem. Your splitting n into 1+1+1…+1 violates condition #1. What you did is invalid.

 

[1MileCrash]

Yeah, it's a very good one though. The error is simply because we are taking the limit as n approaches infinity, making this actually an infinite sum. With any other limit, doing these steps will result in no problems.

 

[krackers]

What is the mathematical reason as to why you can't split the limit when there is an infinite sum?

 

[R136a1]

I don't think the infinite sum is the issue here. There are theorems (like monotone convergence theorem) which thell you that you can split up some infinite sums and limits.

Furthermore, you never actually have an infinite sum in the OP.

What you did is:

\lim_{n\rightarrow +\infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^n \lim_{n\rightarrow +\infty} \frac{1}{n}

The problem is that the sum in the LHS depends on $n$, which is the variable in the limit. You can't just pull something that depends on the variable of the limit outside of the limit. It's the same thing as saying

\lim_{x\rightarrow 0} x = x \lim_{x\rightarrow 0} 1 = x

 

[1MileCrash]

I don't think the infinite sum is the issue here. There are theorems (like monotone convergence theorem) which thell you that you can split up some infinite sums and limits.

Furthermore, you never actually have an infinite sum in the OP.

What you did is:

\lim_{n\rightarrow +\infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^n \lim_{n\rightarrow +\infty} \frac{1}{n}

The problem is that the sum in the LHS depends on $n$, which is the variable in the limit. You can't just pull something that depends on the variable of the limit outside of the limit. It's the same thing as saying

\lim_{x\rightarrow 0} x = x \lim_{x\rightarrow 0} 1 = x

Well, that's a convincing explanation, but you can easily see that all of the steps will cause no problems for any finite limit. That seems to suggest to me that the problem is related to an infinite sum.

 

[R136a1]

Well, that's a convincing explanation, but you can easily see that all of the steps will cause no problems for any finite limit. That seems to suggest to me that the problem is related to an infinite sum.

OK, but there is no actual infinite sum in the OP. However, I do see what you mean. If you change the OP to:

\lim_{n\rightarrow +\infty} \sum_{k=1}^{+\infty} \frac{1}{n} = \sum_{k=1}^{+\infty}\lim_{n\rightarrow +\infty} \frac{1}{n}

then this is also wrong and then the problem is exactly the exchange property for infinite sums.

Some infinite sums can be exchanged though: http://en.wikipedia.org/wiki/Monotone_convergence_theorem#Convergence_of_a_monotone_series But the above does not mean the criteria of the wiki post.

 

[1MileCrash]

OK, but there is no actual infinite sum in the OP.

How is there not an infinite sum?

\lim_{n\to\infty }\Sigma ^{n}_{k=1}f(x)

and

\Sigma ^{\infty}_{k=1}f(x)

are the same thing. It's written the top way in the OP, but that's still an infinite sum.

 

[R136a1]

How is there not an infinite sum?

\lim_{n\to\infty }\Sigma ^{n}_{k=1}f(x)

and

\Sigma ^{\infty}_{k=1}f(x)

are the same thing. It's written the top way in the OP, but that's still an infinite sum.

They are only the same thing if $f(x)$ does not depend on $n$. In this case, "$f(x) = 1/n$" depends on $n$, which is also the upper index of the summation. So we have

\lim_{n\rightarrow +\infty}\sum_{k=1}^{n} f(n)

which is clearly not the same thing as

\sum_{k=1}^{+\infty} f(n)

 

[krackers]

What you did is:

\lim_{n\rightarrow +\infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^n \lim_{n\rightarrow +\infty} \frac{1}{n}

The problem is that the sum in the LHS depends on $n$, which is the variable in the limit. You can't just pull something that depends on the variable of the limit outside of the limit. It's the same thing as saying

\lim_{x\rightarrow 0} x = x \lim_{x\rightarrow 0} 1 = x

If this is the case then why does it hold for any finite limit?

 

[1MileCrash]

They are only the same thing if $f(x)$ does not depend on $n$. In this case, "$f(x) = 1/n$" depends on $n$, which is also the upper index of the summation. So we have

\lim_{n\rightarrow +\infty}\sum_{k=1}^{n} f(n)

which is clearly not the same thing as

\sum_{k=1}^{+\infty} f(n)

Ok, so those two expressions are not the same, but the point is that the first one is still an infinite sum, just not that one. How else can we interpret this expression?

If it is not an infinite sum, it must be a finite sum. Can you tell me how many terms are in it, then?

 

[R136a1]

If this is the case then why does it hold for any finite limit?

Because there is no dependence on $n$ with finite sums. If there is a dependence of $n$, then the result doesn't hold.

So something like

\lim_{n\rightarrow +\infty} \sum_{k=1}^{10} \frac{1}{n} = \sum_{k=1}^{10} \lim_{n\rightarrow +\infty} \frac{1}{n}

because the summation is finite and doesn't depend on $n$.

If you want to know why the equality holds in this case, then you can prove it using analysis techniques and epsilonics. It's a good exercise to see what goes wrong in the infinite case and in the dependence on $n$ case (here the RHS isn't even well-defined).

 

[R136a1]

Ok, so those two expressions are not the same, but the point is that the first one is still an infinite sum, just not that one. How else can we interpret this expression?

If it is not an infinite sum, it must be a finite sum. Can you tell me how many terms are in it, then?

Well, I would argue it's not a sum to begin with. Rather, it's the limit of a sum. So it's the limit of a sum with $n$ terms and where the limit depends on $n$.

 

[krackers]

OK, but there is no actual infinite sum in the OP. However, I do see what you mean. If you change the OP to:

\lim_{n\rightarrow +\infty} \sum_{k=1}^{+\infty} \frac{1}{n} = \sum_{k=1}^{+\infty}\lim_{n\rightarrow +\infty} \frac{1}{n}

then this is also wrong and then the problem is exactly the exchange property for infinite sums.

Some infinite sums can be exchanged though: http://en.wikipedia.org/wiki/Monotone_convergence_theorem#Convergence_of_a_monotone_series But the above does not mean the criteria of the wiki post.

Why is this incorrect? Don't they both evaluate to zero? Also, what are the conditions for which you can exchange the infinite sum and the limit?

 

[1MileCrash]

Well, I would argue it's not a sum to begin with. Rather, it's the limit of a sum. So it's the limit of a sum with $n$ terms and where the limit depends on $n$.

I don't understand how the limit of a sum with $n$ terms as $n$ approaches infinity doesn't qualify as an infinite sum, regardless of how the summands change with the limit value.

 

[R136a1]

I don't understand how the limit of a sum with $n$ terms as $n$ approaches infinity doesn't qualify as an infinite sum, regardless of how the summands change with the limit value.

Because an infinite sum or series is by definition something of the form

\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_k

and not of the form

\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_{n,k}

I will gladly admit that I'm wrong if you can find a reference where the above is called an infinite sum or series.

 

[krackers]

Yeah, it's a very good one though. The error is simply because we are taking the limit as n approaches infinity, making this actually an infinite sum. With any other limit, doing these steps will result in no problems.

Why can you not split the limit in the case of an infinite sum?

 

[R136a1]

Why can you not split the limit in the case of an infinite sum?

Not sure what will satisfy you. You can not split the limit of an infinite sum because you can find counterexamples to it.
Furthermore, the proof of the finite sum-case fails with infinite sums.

 

[1MileCrash]

Because an infinite sum or series is by definition something of the form

\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_k

and not of the form

\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_{n,k}

I will gladly admit that I'm wrong if you can find a reference where the above is called an infinite sum or series.

[strike]But what I'm saying is there is no meaningful distinction between what you and DH are saying. You can choose not to call this an infinite sum if you'd like, but I think we can agree that as the limit approaches infinity, the number of summands approaches infinity.

DH says it's because there are infinite terms as the limit approaches infinity, and we cannot break a limit of sums into a sum of limits if the number of terms is infinite.

You say it's because we cannot switch the sum and the limit operation order because the sum depends on $n$ which is also used in the function, but then this explanation admits that the sum depends on $n$ because the sum is not finite.

It sounds like you and DH are pointing out the same problem with the OP's equation with the caveat that "I don't want to call this an infinite sum." In both cases the underlying problem is that there is an 'infinite sum' appearing, you've offered elaboration on why that causes additional problems in this example besides the simple fact that "we can't break the infinite sum" but it's still because there is an 'infinite sum.'

At the same time, I do understand that we can have a case where the equality fails, your explanation applies, but the sum is not infinite (by any stretch.) But in this case, your explanation applies because the sum is 'infinite.'[/strike]I changed my mind, I don't think that the limit/sum is infinite matters at all, and I think your explanation is right and does not depend on anything.

It is clear that:

\lim_{n\rightarrow m} \sum_{k=1}^n \frac{1}{n}

and

\sum_{k=1}^n \lim_{n\rightarrow m} \frac{1}{n}

are totally different things, and it has nothing to do with infinity and everything to do with the fact that n is used in both the sum and limit.

Thanks

 

[vela]

\lim_{n\rightarrow +\infty} \sum_{k=1}^{+\infty} \frac{1}{n} = \sum_{k=1}^{+\infty}\lim_{n\rightarrow +\infty} \frac{1}{n}

then this is also wrong and then the problem is exactly the exchange property for infinite sums.

Why is this incorrect? Don't they both evaluate to zero? Also, what are the conditions for which you can exchange the infinite sum and the limit?

Nope. The lefthand side is undefined because the sum diverges for all $n$. The way it's written, you have to take the sum first for finite $n$ and then take the limit of that result as $n \to \infty$.

 

[D H]

What you did is:
\lim_{n\rightarrow +\infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^n \lim_{n\rightarrow +\infty} \frac{1}{n}

That isn't what krackers did. Your right hand side is poorly formed, R136a1. The n in your sum is not defined.

Here's what krackers wrote in the OP with my interpretation of what he wrote underneath:\lim_{n\to\infty } 1<br /> = \lim_{n\to\infty }\frac{n}{n}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n}}<br /> _{\displaystyle {\lim_{n \to \infty} \sum_{k=1}^n \frac 1 n}}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}<br /> =0
or
\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n
The problem with this is that exchanging the sum and limit is not a valid operation in this case.

 

[R136a1]

That isn't what krackers did. Your right hand side is poorly formed, R136a1. The n in your sum is not defined.

Yes, I know it's poorly formed, that's exactly my point!

Here's what krackers wrote in the OP with my interpretation of what he wrote underneath:\lim_{n\to\infty } 1<br /> = \lim_{n\to\infty }\frac{n}{n}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n}}<br /> _{\displaystyle {\lim_{n \to \infty} \sum_{k=1}^n \frac 1 n}}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}<br /> =0
or
\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n
The problem with this is that exchanging the sum and limit is not a valid operation in this case.

OK, if that's your interpretation then your explanation is fine by me. But I interpreted it a bit differently:

\lim_{n\to\infty } 1<br /> = \lim_{n\to\infty }\frac{n}{n}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n}}<br /> _{\displaystyle {\lim_{n \to \infty} \sum_{k=1}^n \frac 1 n}}<br /> = \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle n~\text{terms}}<br /> =0

This is what I think the OP means. I know the RHS makes no sense since the $n$ is undefined! But that was exactly my point.

I would like to hear from the OP what exactly the OP means. It would clear up some confusion.

 

[krackers]

Now I'm so confused...

1.

\lim_{n\rightarrow +\infty} \sum_{k=1}^{+\infty} \frac{1}{n}

Nope. The lefthand side is undefined because the sum diverges for all $n$. The way it's written, you have to take the sum first for finite $n$ and then take the limit of that result as $n \to \infty$.

So if I understood correctly, the sum by itself will always diverge to either +\infty or -\infty, and so taking the limit will not affect it?

2.
\sum _{k=1}^{\infty } \lim_{n\to \infty } \frac{1}{n}

How does one evaluate this limit? When you taking the limit of \frac{1}{n} as n tends to \infty, you end up with an infinitely small value, and then when you sum from k=1 to k=\infty, aren't you adding up an infinite amount of infinitesimals? How does that give you back zero (or at least according to mathematica it does).

3.
What's the difference between interpreting the question as this: \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle n~\text{terms}}

and this: \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}

Don't they both essentially have an infinite amount of \lim_{n\to \infty } \, \frac{1}{n} added together?


4.
When you break up the \lim_{n\to \infty } \, \frac{n}{n} into \lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n} there need to be n 1's as only then will 1\cdot n\; =\; n. However, since n\to \infty, isn't this the same as having \lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{\infty \text{ times}}}{n}?

5.

So if you choose to interpret it as \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle n~\text{terms}}

(which I don't know if it is the right way to interpret it)

The issue is that the n in "n terms" has been taken out of the context of the limit as as such n is undefined?

But isn't it still obvious that the number of terms is tending to \infty?

6.

If you choose to interpret it as

\underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}

(which, again, I don't know if it is the right way to interpret it)

the issue is that you cannot split a limit of infinite terms into an infinite sum of limits?

I still do not quite understand why this cannot be done

Not sure what will satisfy you. You can not split the limit of an infinite sum because you can find counterexamples to it.
Furthermore, the proof of the finite sum-case fails with infinite sums.

To me, it doesn't seem like enough of a justification that the only reason is that there are counterexamples. Isn't there any mathematical reason as to why it cannot be done? Further, it can be proved that the limit law holds for any n finite number of terms, so what prevents you from just continually incrementing n, essentially making it go towards infinity?

 

[D H]

Now I'm so confused...

1.


So if I understood correctly, the sum by itself will always diverge to either +\infty or -\infty, and so taking the limit will not affect it?

Remember what a limit is. $\lim_{n \to \infty} f(n)$ is not about the behavior of f(n) at infinity. f(∞) doesn't make sense. The limit is about the behavior of f(n) as n grows ever larger but remains finite. If every f(n) is infinite, so is the limit (better: The limit doesn't exist).

2.
\sum _{k=1}^{\infty } \lim_{n\to \infty } \frac{1}{n}

How does one evaluate this limit? When you taking the limit of \frac{1}{n} as n tends to \infty, you end up with an infinitely small value, and then when you sum from k=1 to k=\infty, aren't you adding up an infinite amount of infinitesimals? How does that give you back zero (or at least according to mathematica it does).

This is (informally) the sum of an infinite number of zeros. It's zero.

3.
What's the difference between interpreting the question as this: \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle n~\text{terms}}

and this: \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}

Don't they both essentially have an infinite amount of \lim_{n\to \infty } \, \frac{1}{n} added together?

No. The first is ill-formed. What is n in that "n terms"? The second, as noted, is zero.

4.
When you break up the \lim_{n\to \infty } \, \frac{n}{n} into \lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n} there need to be n 1's as only then will 1\cdot n\; =\; n. However, since n\to \infty, isn't this the same as having \lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{\infty \text{ times}}}{n}?

No.

5.

So if you choose to interpret it as \underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle n~\text{terms}}

(which I don't know if it is the right way to interpret it)

The issue is that the n in "n terms" has been taken out of the context of the limit as as such n is undefined?

Exactly.

But isn't it still obvious that the number of terms is tending to \infty?

Obviously not.

6.

If you choose to interpret it as

\underbrace<br /> {\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots}<br /> _{\displaystyle {\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac 1 n}}

(which, again, I don't know if it is the right way to interpret it)

the issue is that you cannot split a limit of infinite terms into an infinite sum of limits?

I still do not quite understand why this cannot be done

For the simple reason that you get nonsense. Anytime you have found that 1=0 you have violated some rule.

Various things you thought were always true are not necessarily so when dealing with infinite sums and limits. For example, a+b+c=b+c+a=c+a+b=… Rearranging the terms in a sum of a finite number of terms doesn't change the sum. This is not necessarily true for infinite series. If a series is conditionally convergent but not absolutely convergent, you can rearrange the terms to get any number you want.

 

[1MileCrash]

How does one evaluate this limit? When you taking the limit of \frac{1}{n} as n tends to \infty, you end up with an infinitely small value

No, you end up with exactly zero. I never understand why people want to call that an "infinitely small value." The limit is zero. So the infinite sum is a sum of zeros. So its zero.

To me, it doesn't seem like enough of a justification that the only reason is that there are counterexamples. Isn't there any mathematical reason as to why it cannot be done? Further, it can be proved that the limit law holds for any n finite number of terms, so what prevents you from just continually incrementing n, essentially making it go towards infinity?

I maintain that R136a1's reason is precisely why the equality fails. I look at his explanation in the following way: with n being involved in both the sum and the limit, you could look at the summation of some function/object "f" as being a function of n (the upper limit of summation), and the limit of some function/object "f" as being a function of n (the variable to approach the limit value). You could then look at the equation as a whole as being a composition of those two functions of n and f, and in the "problem move" we have reversed the composition order, and compositions of functions generally aren't commutative except in a few cases (I don't know them).

Or to say differently, let
L(f,n) = \lim_{n\rightarrow \infty} f

and

S(f,n) = \sum_{i=1}^n f

We can then say that
(L \circ S)(f,n) = \lim_{n\rightarrow \infty} \sum_{i=1}^n f
and
(S \circ C)(f,n) = \sum_{i=1}^n \lim_{n\rightarrow \infty} f

And the composition of functions is almost never commutative.

This may be nonsense of course, but I have convinced myself that the problem only occurs whenever "n" is involved in both the sum and limit, and this is how I reason why. If n is not involved in both the sum and limit, it is no longer a composition of two functions in the normal sense.We can see that we still have hickups when the limit does not approach infinity, suggesting that the problem is that both the limit and sum involve the same term (n) and thus cannot be inverted, and has nothing to do with infinity.

Observe that

\lim_{n\rightarrow q} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^q \frac{1}{q} = 1

but

\sum_{k=1}^n \lim_{n\rightarrow q} \frac{1}{n} = \sum_{k=1}^n \frac{1}{q} = \frac{n}{q}

Which means the inverted composition is equal to the original only if n = q, but that just makes the limit an identity function, which I'm sure is one of the few cases when a composition is commutative (obviously.)

 

[johnqwertyful]

When you have a result that is incorrect, something you did in your computations was incorrect. When asking why you can't do something in your computations, it is because it leads to incorrect answers; namely, your answer.

You're asking why you can't perform a certain step, and it is because your conclusion is wrong. Sometimes it's helpful to figure out exactly what breaks down. But rather than ask why you can't perform a certain step, ask yourself why you think you would be able to perform that step. Why would you think it's okay to go from step 3 to 4.

 

[R136a1]

aren't you adding up an infinite amount of infinitesimals? How does that give you back zero (or at least according to mathematica it does).

DH, 1MileCrash and johnqwertyful have given you excellent answers already. But I wish to address this.

Infinitesimals do not exist in real analysis. So $\lim_{n\rightarrow +\infty}\frac{1}{n} = 0$ and it is exactly zero, not some infinitesimal value larger than zero.

I know you can make sense of infinitesimals and nonstandard analysis, but then you're changing the context of your question. As you asked the question, you asked about the real analysis that more 99% of the mathematicians work with and in that world, the limit above is exactly $0$.

 

[R136a1]

To me, it doesn't seem like enough of a justification that the only reason is that there are counterexamples. Isn't there any mathematical reason as to why it cannot be done? Further, it can be proved that the limit law holds for any n finite number of terms, so what prevents you from just continually incrementing n, essentially making it go towards infinity?

From a rigorous and mathematical point of view, the existence of counterexamples alone is enough justification that it cannot be done.

I understand why this might not be an intuitive justification however. But giving you an intuitive reason for why it cannot be done is difficult. In fact, it depends a lot on you! Try to prove why it actually can be done and see where the proof breaks down.

 

[Tombaitsfas]

R1 is right