1. ## Series.

Hello,

I have two problems with series first $\displaystyle \sum^{\infty }_{n=1}\frac{(-1)^nn^2}{n^3+1}$

so I will like to know of this is absolute convergent here for I use two test.

First $\displaystyle \frac{n^2}{n^3+1} =^?0$ and I think this is oké $\displaystyle \frac{n^2(1)}{n^2(n+\frac{1}{n^2})}= \frac{1}{n+ \frac{1}{n^2}}=0$

then the second test $\displaystyle \frac{n^2}{n^3+1}<\frac{(n+1)^2}{(n+1)^3+1}$ but the dominator will be bigger then te nominator?

Or not?

Second problem $\displaystyle \sum^{\propto}_{n=1}\frac{n}{2n+1}$

I will proof that this is convergent $\displaystyle \frac{\frac{(n+1)}{2n+3}}{\frac{n}{2n+1}}=\frac{n+ 1}{2n+3}\frac{2n+1}{n}$

if I calculated this then I get $\displaystyle \frac{5}{2}$ thus not convergent but it must be.

Who can help me out? Greets.

2. Hello, Bert!

I have an "eyeball" approach to #1 . . . which may not be acceptable.

$\displaystyle 1)\;\;\sum^{\infty}_{n=1}\frac{(-1)^nn^2}{n^3+1}$
Is this absolutely convergent?
For large $\displaystyle n:\;\;\frac{n^2}{n^3 + 1} \approx \frac{1}{n}$ . . . which is the divergent Harmonic Series.

Hence the given series also diverges.

$\displaystyle 2)\;\;\sum^{\infty}_{n=1}\frac{n}{2n+1}$
$\displaystyle \lim_{n\to\infty} a_n\;=\;\lim_{n\to\infty}\frac{n}{2n+1}$

Divide top and bottom by $\displaystyle n\;\;\lim_{n\to\infty}\frac{1}{2 + \frac{1}{n}} \;= \;\frac{1}{2 + 0} \;= \;\frac{1}{2}$

Since the $\displaystyle n^{th}$ term does not approach $\displaystyle 0$, the series diverges.

3. but normally you can test so harmonica serie with two test or not?

4. For the first one, You can show absolute divergence by using the integral test.

Consider:

$\displaystyle \int_1^\infty \frac{x^2}{x^3+1}\,dx$

Integrate by substitution to get:

$\displaystyle \lim_{b\to\infty} \left( \frac{1}{3}\ln\left|b^3+1\right| - \frac{1}{3}\ln 2 \right) = \infty$

Since this integral diverges, so does $\displaystyle \sum_{n=1}^\infty \frac{n^2}{n^3+1}$ (which is the absolute value of the terms of the alternating sum).