MHB Converge conditionally or converge absolutely

[shamieh]

I'm confused. I know that $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$ but I don't know what it does when it is alternating...

I'm given this $$\sum^{\infty}_{n= 1} \frac{(-1)^n}{n^2} $$ and told to find out if it conditionally converges, diverges or converges absolutely.Can I use lopitals to get 1/n and then use the AST and say 0 <= 1/n+1 <= 1/n and lim n-> infinity = 0 therefore by AST the series converges?

 

[Prove It]

I'm confused. I know that $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$ but I don't know what it does when it is alternating...

I'm given this $$\sum^{\infty}_{n= 1} \frac{(-1)^n}{n^2} $$ and told to find out if it conditionally converges, diverges or converges absolutely.Can I use lopitals to get 1/n and then use the AST and say 0 <= 1/n+1 <= 1/n and lim n-> infinity = 0 therefore by AST the series converges?

Since you already know that $\displaystyle \begin{align*} \sum \left| \frac{ (-1)^n }{n^2} \right| = \sum \frac{1}{n^2} \end{align*}$ is convergent, the series is obviously absolutely convergent...

 

[shamieh]

Wow I'm making these way harder than they need to be. I need to get better at inspection. But couldn't you also use the AST test since you really just have a $$\frac{1}{n^2}$$ which is a number over infinity, which = 0 ? Also couldn't you just look at $$1/n^2$$ using the p series test as well to see that $$p > 1$$ so converges?

 

[Prove It]

Yes LAST can be used to determine the convergence of this series as well.

HOWEVER, your question is asking you to determine if the series is ABSOLUTELY convergent or CONDITIONALLY convergent. That means you need to test for absolute convergence first. You would go for LAST if you couldn't show the series is absolutely convergent.