Absolute convergence2

I have to test the series for absolute and conditional convergence

$\sum_{n=2}^{\infty}$ $\frac{(-1)^{n-1}}{n^2+(-1)^n}$

$Notes :$ For absolute convergence def. I have $\vert \frac{(-1)^{n-1}}{n^2+(-1)^n} \vert$ If this convergence then the original series should be convergence.

$proof:$ $\vert \frac{(-1)^{n-1}}{n^2+(-1)^n} \vert$ = $\frac {1}{n^2+1}$ $\lt$ $\frac{1}{n}$

as this sequence decreasing to $0$ is diverges. Therefore the series is not absolute convergence but it's is conditional convergence by alternating series test.

Any help would be grateful ;)

• $\frac{1}{n^2+1}<\frac{1}{n}...$ – Cyclohexanol. Dec 7 '14 at 2:27
• The first inequality in the line starting with "proof" is incorrect. Check this for $n = 1$! The second inequality is correct, but the conclusion then is incorrect. – Hans Engler Dec 7 '14 at 2:31
• Try $\left|\frac{(-1)^{n-1}}{n^2+(-1)^n}\right| = \frac {1}{n^2\pm 1} < \frac{1}{(n-1)^2}$. You know that $\sum_{n=2}^\infty\frac{1}{(n-1)^2} = \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}$ converges. – Wood Dec 7 '14 at 2:36

Considering absolute convergence, the partial sum of the series is

$$\sum_{n=2}^{m}|x_n| = \sum_{n \,\text{even}}\frac{1}{n^2+1}+ \sum_{n \,\text{odd}}\frac{1}{n^2-1}.$$

You can show by comparison that both series on the RHS converge.

• thank you..let me see if i can do it – tommy Dec 7 '14 at 2:39
• i try to do odd but i got $\frac{1}{n^2-1}$$\lt$$\frac{1}{(n+1)(n-1)}$ then what – tommy Dec 7 '14 at 2:53
• The other posts answer this more efficiently. But basically $n^2-1 > (n-1)^2$ if $n > 1$ so you can compare with $1/n^2$ for either even or odd – RRL Dec 7 '14 at 3:08

You may write \begin{align} \sum_{n=2}^{\infty}\left|\frac{(-1)^{n-1}}{n^2+(-1)^n}\right| &\le\sum_{n=2}^{\infty}\frac{1}{n^2-1}<\sum_{n=2}^{\infty}\frac{1}{(n-1)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}<+\infty \end{align} and your series is absolutely convergent.

Even simpler:

$$\begin{array}\\ \sum_{n=2}^m |x_n| &=\sum_{n=2}^m\frac{1}{n^2+(-1)^n}\\ &\le\sum_{n=2}^m\frac{1}{n^2-1}\\ &<\sum_{n=2}^m\frac{1}{n^2-n}\\ &=\sum_{n=2}^m\left(\frac1{n-1}-\frac1{n}\right)\\ &=1-\frac1{m} < 1\\ \end{array}$$

• why the even be negative $n^2-1$ – tommy Dec 7 '14 at 2:51
• I wanted to get an inequality that was true for all $n$ so I didn't have to handle even and odd separately. – marty cohen Dec 7 '14 at 2:54
• so that it? what about if its conditional convergence? – tommy Dec 7 '14 at 2:55
• If it converges absolutely, then it converges conditionally. But not the converse - standard example is $\sum \frac{(-1)^n}{n}$. – marty cohen Dec 7 '14 at 2:57
• About $\sum_{n=2}^m\left(\frac1{n-1}-\frac1{n}\right)$: en.wikipedia.org/wiki/Telescoping_series – Wood Dec 7 '14 at 3:00