Show That The Sequence Converges Mistake?

[mmmboh]

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

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/n^1/2, and the summation of 1 from k=1 to n is just n, right?

so that gives n*1/n^1/2-2n^1/2= n^1/2-2n^1/2=-n^1/2...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.

 

[Office_Shredder]

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

 

[mmmboh]

Ok so turns out there was a mistake, the 1/n^1/2 in the summation should be 1/k^1/2. I'm not sure how to solve it though. I tried doing it by showing the sequence is cauchy, and I got a_m+n-a_n (for some m in the reals) = $$\Sigma$$1/k^1/2 from k=n+1 to k=n+m (the other 1/k^1/2'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?...

 

[micromass]

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

 

[micromass]

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.

 

[mmmboh]

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

 

[micromass]

Try to prove first that the sequence is monotonically increasing.

 

[mmmboh]

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

 

[micromass]

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

 

[mmmboh]

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

 

[micromass]

$$n-\sqrt{n^2+n}>-\frac{1}{2}$$
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...

EDIT: made a stupid mistake

 

[mmmboh]

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!

 

[micromass]

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?

 

[mmmboh]

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).

 

[mmmboh]

Eek induction isn't working :S.

 

[micromass]

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 positive) 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.

 

[mmmboh]

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

 

[micromass]

What inequality do you mean? This one?

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

That's by the induction hypothesis.

 

[mmmboh]

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

 

[micromass]

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...

 

[mmmboh]

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?

 

[micromass]

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.

 

[mmmboh]

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!

 

[micromass]