Will This Mathematical Series Converge?

[zonk]

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.

 

[micromass]

Hi zonk!

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

What book are you using anyway?

 

[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.

 

[micromass]

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

 

[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.

 

[micromass]

Well,

4n+1\leq n^2

For large n. Thus

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

and you can factor the 2 out.

 

[zonk]

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

 

[Ray Vickson]

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