# Convergence of a Series

1. Jul 15, 2011

### zonk

1. The problem statement, all variables and given/known data

Test for convergence or divergence. Give a reason for your decision.

2. Relevant equations

$\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n(n + 1)}$

3. The attempt at a solution

I've tried to compare it to the series $\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n^2}$ and show the latter converges. I have no idea how to show this. Although the limit of the sequence approaches 0 as n goes to infinite, that is not enough to guarantee convergence. The book says it converges.

Last edited: Jul 15, 2011
2. Jul 15, 2011

### micromass

Staff Emeritus
Hi zonk!

Do you know the Cauchy condensation test? The integral test? These two could prove useful here.

What book are you using anyway?

3. Jul 15, 2011

### zonk

It's Apostol volume 1. I think Cauchy's test is a section or two after this. Yes he did teach the integral test in this section.

4. Jul 15, 2011

### micromass

Staff Emeritus
OK, can you bring the series to something of the form

$$\sum{\frac{\log(n)}{n^{2-\frac{1}{2}}}}$$

this is suitable for the integral test

5. Jul 15, 2011

### zonk

I can reduce it to:

$\sum_{i=1}^{\infty} \frac{\sqrt{2} \log{(4n + 1)}}{n^{(2 - \frac{1}{2})}}$ and can factor sqrt(2) out, but I don't see how you can get the log argument that way.

Last edited: Jul 15, 2011
6. Jul 15, 2011

### micromass

Staff Emeritus
Well,

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

For large n. Thus

$$\log(4n+1)\leq 2\log(n)$$

and you can factor the 2 out.

7. Jul 15, 2011

### zonk

Oh, thank you so much, I would have never figured this out without those hints.

8. Jul 15, 2011

### Ray Vickson

Let t(n) = nth term above. You could try to get a simple upper bound on t(n): sqrt(2n-1) < sqrt(2n), log(4n-1) < log(4n) and n(n+1) > n^2. Thus, t(n) < sqrt(2n)*log(4n)/n^2, which is of the form c*log(n)/n^(3/2). Convergence of sum log(n)/n^(3/2) is easier to show, and that implies convergence of sum t(n) [why?]

RGV