# Show That The Sequence Converges Mistake?

• mmmboh
In summary, the conversation revolves around a series that appears to diverge to negative infinity. However, it is later discovered that there was a mistake in the equation and the sequence actually converges to negative infinity. Various methods are discussed for proving the convergence of the sequence and for finding a lower bound. One method involves using integrals to find an inequality, which is then used to prove the monotonicity of the sequence. Ultimately, it is shown that the sequence is bounded below by -2. The use of integrals is debated as a valid method of finding the inequality.

#### mmmboh

[PLAIN]http://img408.imageshack.us/img408/6862/mistakeg.jpg [Broken]

Am I reading this wrong, or does this sequence not converge?
There is no k dependence in the summation, so we can take out the 1/n1/2, and the summation of 1 from k=1 to n is just n, right?

so that gives n*1/n1/2-2n1/2= n1/2-2n1/2=-n1/2...and this sequence diverges to negative infinity...

Am I misunderstanding something? Or does my teacher consider this sequence to be "converging" to negative infinity, which seems strange.

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mmmboh said:
Or does my teacher consider this sequence to be "converging" to negative infinity, which seems strange.

There's a standard definition for converging to negative infinity which probably agrees with your intuition, but if you haven't covered it in class then "does not exist" is the right thing to write down

Ok so turns out there was a mistake, the 1/n1/2 in the summation should be 1/k1/2. I'm not sure how to solve it though. I tried doing it by showing the sequence is cauchy, and I got am+n-an (for some m in the reals) = $$\Sigma$$1/k1/2 from k=n+1 to k=n+m (the other 1/k1/2's cancel out)...I'm not sure how I am suppose to choose my n though so that the sequence is less than a given $$\epsilon$$...any help? Is this even a good method?...

Well, you could prove that the sequence is strictly increasing/decreasing and that it's bounded. Not sure if that would work tho...

Ok, I would use the following not-so-hard-to-prove inequalities:

$$\sum_{k=2}^n{\frac{1}{\sqrt{k}}}\leq \int_1^n{\frac{1}{\sqrt{x}}dx}\leq\sum_{k=1}^{n-1}{\frac{1}{\sqrt{k}}}$$

Then you can change the series by values you can actually work with.

Hm it's for my analysis class, we haven't talked about integration yet. I just realized I wrote out (an+1)-(an) wrong though. It's actually
1/(n+1)1/2-2((n+1)1/2-n1/2)...so I think it's monotone down but I have to show it, and then find a lower bound I guess...

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Try to prove first that the sequence is monotonically increasing.

It's not monotone increasing I wrote it out wrong, see above...I think it's monotone decreasing though.

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Strange, I thought I have proven that it was increasing. Must have been something wrong with the proof then I'll recheck...

Ok I figured out if I can show n-(n^2+n)1/2>-1/2, then it is monotone decreasing.

$$n-\sqrt{n^2+n}>-\frac{1}{2}$$
if and only if
$$2n+1 > 2\sqrt{n^2+n}$$
Squaring both sides is legal since all terms are positive. So this is true iff

$$4n^2+4n+1>4n^2+4n$$

This is true iff

$$1>0$$

This is true, so I guess its decreasing...

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No it is true, you just forgot to square the left side, 2n+1...the inequality should be (2n+1)2=4n2+4n+1>4n2+4n, which is true. I figured it out too, but I like yours a little better. Thanks...now I just have to find a way to show it is bounded.

Edit: Oh you saw the mistake!

What I think to be true is:

$$2\sqrt{n+1}-2\sqrt{n}-2\leq \sum_{k=1}^n{\frac{1}{\sqrt{k}}}-2\sqrt{n}$$

This is easy to check with integrals. But I don't know how to check it without them. Induction maybe?

Hm I'm not sure yet. What would that accomplish though, are you trying to bound it below by -2? (I think the summation should be from 1 to n+1 if so, but anyway).

Eek induction isn't working :S.

Ok, spare me if I made some mistakes, but this seems to be the proof of the inequality:

$$2\sqrt{n+1}-2\sqrt{n}-2\leq \sum_{k=1}^n{\frac{1}{\sqrt{k}}-2\sqrt{n}$$

Obviously, it suffices to show that

$$2\sqrt{n+1}-2\leq \sum_{k=1}^n{\frac{1}{\sqrt{k}}$$

Let's proceed by induction. For n=1, this is easy. So let's assume the inequality is true for n-1. Then we have that

$$\sum_{k=1}^n{\frac{1}{\sqrt{k}}=\sum_{k=1}^{n-1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n}} \geq 2\sqrt{n}-2+\frac{1}{\sqrt{n}}$$

We need to show that

$$2\sqrt{n}-2+\frac{1}{\sqrt{n}}\geq 2\sqrt{n+1}-2$$

Multiplying by $$\sqrt{n}$$ (this is positive) yields that this is true iff

$$2n+1\geq 2\sqrt{n(n+1)}$$

Squaring both sides (which are postive) gives that this is true iff

$$4n^2+4n+1\geq 4n^2+4n$$

But this is obviously true. So the original inequality has to be true. So

$$2\sqrt{n+1}-2\sqrt{n}-2\leq \sum_{k=1}^n{\frac{1}{\sqrt{k}}-2\sqrt{n}$$

But the sequence $$2\sqrt{n+1}-2\sqrt{n}-2$$ is a decreasing function, whose limit is -2. So we have that the sequence

$$\sum_{k=1}^n{\frac{1}{\sqrt{k}}-2\sqrt{n}$$

is bounded below by -2.

Ya looks good thanks. Just one thing, I am unsure of where the first inequality came from?

What inequality do you mean? This one?

micromass said:
$$\sum_{k=1}^n{\frac{1}{\sqrt{k}}=\sum_{k=1}^{n-1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n}} \geq 2\sqrt{n}-2+\frac{1}{\sqrt{n}}$$

That's by the induction hypothesis.

No this one
$$2\sqrt{n+1}-2\sqrt{n}-2\leq \sum_{k=1}^n{\frac{1}{\sqrt{k}}-2\sqrt{n}$$

Ow, where I got it from. Well, you will not like it, but it's a standard way of finding inequalities.

If $$f:\mathbb{R}^+\rightarrow \mathbb{R}^+$$ is a decreasing function, then

$$\int_1^{n+1}{f(x)dx}\leq \sum_{k=0}^n{f(k)}$$

So performing the integral with $$f(x)=\frac{1}{\sqrt{x}}$$ will give you the inequality

$$2\sqrt{n+1}-2\leq \sum_{k=0}^n{f(k)}$$

I realize your not supposed to use integral, but I see no other way of finding that inequality. But at least the proof of the inequality is integral-free...

But then if that inequality just proves it is decreasing, then how does it show it is bounded below by -2?

Edit: Oh wait I see. But is there a way I can justify this inequality without saying I found it using an integral?

I'm sorry, but the only way I know of finding this inequality is by integrating... Maybe you can say it way by trial-and-error.

Yeah actually, I don't think I need to justify where it came from. If I show it is true then my sequence must be bounded below by -2, proof is proof. Thanks a lot!

## What is the definition of convergence in a sequence?

Convergence in a sequence means that as the sequence progresses, the terms get closer and closer to a single number, known as the limit. In other words, the sequence approaches a specific value as the number of terms increases.

## What are the common mistakes made when showing that a sequence converges?

One common mistake is assuming that a sequence converges without properly proving it. Another mistake is using incorrect or incomplete methods, such as using the limit of the sequence as the limit of the series. It is also important to consider the behavior of the sequence as the number of terms approaches infinity, as this can impact the convergence.

## Can a sequence have multiple limits?

No, a sequence can only have one limit. However, it is possible for a sequence to have no limit, or to approach different limits from different directions.

## What are some methods for showing that a sequence converges?

One method is to show that the sequence is monotonic (either increasing or decreasing) and bounded. Another method is to use the definition of convergence, and show that for any given error tolerance, there exists a term in the sequence that is within that tolerance of the limit.

## How can I avoid making mistakes when showing that a sequence converges?

To avoid mistakes, it is important to carefully follow the steps and methods for proving convergence. Double check your calculations and make sure to consider the behavior of the sequence as the number of terms approaches infinity. It can also be helpful to have a peer or mentor review your work for any errors or oversights.