# Thread: convergence problems

1. ## convergence problems

Hey all I got two convergence problems I am stuck on

1) Determine whether 1/(x((lnx)^2)) from 2 to infinity converges or diverges. I've thought about the integral test and the comparison test but I'm still stuck.

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2) Determine whether (-1)^(n-1)*(n/((n^2)+1)) from 1 to infinity converges or diverges and if it converges what's its limit. I tried direct comparison but that broke down at n=4 so I'm not sure what to do know.

Thanks for any help

2. The integral test should work for the first...

Note that $\displaystyle \displaystyle \frac{1}{n\ln^2{n}}$ is a decreasing function for $\displaystyle \displaystyle n \geq 2$, so

$\displaystyle \displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n\ln^2{n}}} \leq \int_2^{\infty}{\frac{1}{x\ln^2{x}}\,dx}$.

Let $\displaystyle \displaystyle u = \ln{x}$ so that $\displaystyle \displaystyle \frac{du}{dx} = \frac{1}{x}$.

When $\displaystyle \displaystyle x = 2, u = \ln{2}$ and when $\displaystyle \displaystyle x \to \infty, u \to \infty$, so

$\displaystyle \displaystyle \int_2^{\infty}{\frac{1}{x\ln^2{x}}\,dx} = \int_{\ln{2}}^{\infty}{u^{-2}\,du}$

$\displaystyle \displaystyle = \lim_{\epsilon \to \infty}\left[-u^{-1}\right]_{\ln{2}}^{\epsilon}$

$\displaystyle \displaystyle = \lim_{\epsilon \to \infty}\left(\frac{1}{\ln{2}} - \frac{1}{\epsilon}\right)$

$\displaystyle \displaystyle = \frac{1}{\ln{2}} - 0$

$\displaystyle \displaystyle = \frac{1}{\ln{2}}$.

Therefore $\displaystyle \displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n\ln^2{n}}} \leq \frac{1}{\ln{2}}$ and thus the series is convergent.