Find if this sequence has a limit

Question

Does this sequence (with index $n$) has got a limit? If yes - find it: $\sum_{k=0}^{n} \frac{(-1)^k}{2k+1}(\frac{1}{3})^k$

Generally, I know that things like this need to be done with Riemann sums - but I do not know how to deal with that example, especially with that $(-1)^k$. Any ideas?

Thanks a lot in advance for your help!

Answer 1

First of all, whenever you have a $(-1)^k$ disturbing you, it can simply mean you have to take the value in $-\frac{1}{3}$ instead (it is not harder at all)

As for the Riemann sum, have you tried $\arctan$ ? I just have like a feeling it might help...

If you are still unhappy, try

$x \rightarrow \frac{\arctan(\sqrt{x})}{\sqrt{x}}$

Answer 2

You do not need Riemann sums:

Note that $\sum_{k=0}^{n} \frac{(-1)^k}{2k+1}\left ( \frac{1}{3} \right )^k=\sqrt{3}\sum_{k=0}^{n} \frac{(-1)^k}{2k+1}\left ( \frac{1}{\sqrt{3}} \right )^{2k+1}$.

Now consider the series $\sum_{k=0}^{n }(-1)^{k}x^{2k}=\sum_{k=0}^{n }(-1)^{k}\left ( x^{2} \right )^{k}$.