1. ## Alternating Series

I have trouble with alternating series.
I have to test the series for convergence or divergence.
Any help would be appreciated
Thanks!

2. Originally Posted by BACONATOR
I have trouble with alternating series.
I have to test the series for convergence or divergence.
Any help would be appreciated
Thanks!
Hello,

Remember that for an alternating series to converge $\displaystyle \sum_{n} (-1)^n \ a_n$, you have to check that :

- $\displaystyle \lim_{n \to \infty} a_n=0$
- $\displaystyle a_n$ is a decreasing sequence

But here, one can see that when n approaches high values, +2 can be neglected.

Thus the general term is equivalent to $\displaystyle (-1)^n \frac{n}{\sqrt{n^3+2}} \sim (-1)^n \frac{n}{\sqrt{n^3}}=(-1)^n \frac{n}{n^{3/2}}=(-1)^n \frac{1}{n^{1/2}}$, which is an alternating Riemann series that converges.

3. Hi
Originally Posted by Moo
Thus the general term is equivalent to $\displaystyle (-1)^n \frac{n}{\sqrt{n^3+2}} \sim (-1)^n \frac{n}{\sqrt{n^3}}=(-1)^n \frac{n}{n^{3/2}}=(-1)^n \frac{1}{n^{1/2}}$, which is an alternating Riemann series that converges.
You can't use approximations ($\displaystyle \sim$) in this case since the sign of $\displaystyle (-1)^n\frac{n}{\sqrt{n^3+2}}$ changes : The theorem you're using states that if $\displaystyle {\color{red}a_n,\,b_n>0}$, if $\displaystyle a_n\sim b_n$ and if $\displaystyle \sum a_n$ converges then $\displaystyle \sum b_n$ converges.

4. Originally Posted by BACONATOR
I have trouble with alternating series.
I have to test the series for convergence or divergence.
Any help would be appreciated
Thanks!
We can see that this satisfies the two conditions of the converging series

$\displaystyle \sum_{n=0}^{\infty}(-1)^na_n$

Firstly

$\displaystyle \exists{N}\backepsilon\forall{n>N}\quad{a_{n+1}<a_ n}$

Furthermore

$\displaystyle \lim_{n\to\infty}a_n=0$

$\displaystyle \Rightarrow\quad\sum_{n=0}^{\infty}(-1)^n\frac{n}{\sqrt{n^3+2}}~~\text{converges}$