Integral Test can it be used here

Question

Can we apply integral test on

Series with $(1-n)/2n^2$

First thing I did was to take $-1/2$ outside

So we have

$$(-1/2)\sum((n-1)/n^2)$$

I'm not sure how to follow up later although I know for sure this series will diverge

Can someone help to prove its decreasing too

Answer 1

Yes, you can apply it:$$\int_1^\infty\frac{x-1}{x^2}\,\mathrm dx=\lim_{M\to\infty}\log(M)+\frac1M-1=\infty$$and therefore the series $\sum_{n=1}^\infty\frac{n-1}{n^2}$ diverges.

Answer 2

The very simple way uses asymptotic equivalents:

$1-n\sim_\infty -n$, so $\dfrac{1-n}{2n^2}\sim_\infty \dfrac{-n}{2n^2}=-\dfrac 1{2n}$, and the latter diverges.

This uses the following general result from Asymptotic analysis:

If two series $\sum a_n$ and \sum b_n$ have equivalent general terms of constant sign, the both converge or both diverge

Answer 3

You do not need integral test for this series. Integral test is messy. You can just look at $\frac{n - 1}{n^2}$ you will get $\frac{1}{n} - \frac{1}{n^2}$. Based on p series theorem, if $\frac{1}{n^p}$ where p $\leq 1$ , then the series is divergent. thus, the series you have is divergent