# Convergence of a Series

## Homework Statement

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

## Homework Equations

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

## 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.

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

What book are you using anyway?

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.

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

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.

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Well,

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

For large n. Thus

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

and you can factor the 2 out.

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

Ray Vickson
Homework Helper
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## Homework Statement

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

## Homework Equations

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

## 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.
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