# Series.

• Jun 24th 2006, 09:21 AM
Bert
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.
• Jun 24th 2006, 10:38 AM
Soroban
Hello, Bert!

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

Quote:

$\displaystyle 1)\;\;\sum^{\infty}_{n=1}\frac{(-1)^nn^2}{n^3+1}$
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.

Quote:

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

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.
• Jun 24th 2006, 10:53 AM
Bert
but normally you can test so harmonica serie with two test or not?
• Jun 24th 2006, 10:56 AM
Soltras
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).