# Thread: Prove convergence and find the sum

1. ## Prove convergence and find the sum

$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n}$

I can figure out how to prove this series converges, but I cannot figure out how to find its sum.

I know this is the answer:
$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n} = \frac{log(2)}{2}$

But how?

I got a hint to use this: $tan^nu$ $du = \frac{1}{n-1}tan^{n-1}u-\int{tan^{n-2}}du$

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Proof it converges:

$b_n =\frac{1}{2n}$
$b_{n+1}=\frac{-1}{2n^2}$

Since $n=1$, $n$ is always positive,

$b_n > b_{n+1}$

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A hint to get me started would be awesome.

2. Originally Posted by Cursed
$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n}$

I can figure out how to prove this series converges, but I cannot figure out how to find its sum.

I know this is the answer:
$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n} = \frac{log(2)}{2}$

But how?

I got a hint to use this: $tan^nu$ $du = \frac{1}{n-1}tan^{n-1}u-\int{tan^{n-2}}du$

________________________________________________
Proof it converges:

$b_n =\frac{1}{2n}$
$b_{n+1}=\frac{-1}{2n^2}$

Since $n=1$, $n$ is always positive,

$b_n > b_{n+1}$

________________________________________________

A hint to get me started would be awesome.
That hint is absolutely stupid.

$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\implies \int\frac{dx}{1+x}=\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n}$ ignoring all of the justification since I don't think you care about that.

3. Thanks.

But how would I go about using that so-called "hint"? I figure the squeeze lemma and integral test would apply -- I just don't know how exactly I'd use them.