Find $\left( \sum_{n=0}^{\infty} {(-1)^nx^{n+1} \over n + 1} \right)\left( \sum_{n=0}^{\infty} (-1)^nx^n \right)$ I am looking for the simplest form of
$$\left( \sum_{n=0}^{\infty} {(-1)^nx^{n+1} \over n + 1} \right)\left( \sum_{n=0}^{\infty} (-1)^nx^n \right)$$
We can use the Cauchy product
$$\sum_{i=0}^{n}a_i b_{n-i}=\sum_{i=0}^{n}{(-1)^ix^{i+1} \over i + 1}(-1)^{n-i} x^{n-i}=\sum_{i=0}^{n}{(-1)^n x^{n+1} \over i + 1}$$
I suspect this can be written down in a trigonometric form involving the sine or cosine. How can I get from this expression to the trigonometric form?
 A: Hints:
Remember that for $\;|x|<1\;$ 
$$\frac1{1+x}=\sum_{n=0}^\infty (-1)^nx^n\implies\log(1+x)=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}$$
A: One may recall that
$$
 \sum_{n=0}^{\infty} (-1)^nx^{n}  =\frac1{1+x},\quad |x|<1,
$$ and that
$$
 \sum_{n=0}^{\infty} {(-1)^nx^{n+1} \over n + 1} =\ln(1+x),\quad |x|<1,
$$ then the simplest form of 
$$
\left( \sum_{n=0}^{\infty} {(-1)^nx^{n+1} \over n + 1} \right)\left( \sum_{n=0}^{\infty} (-1)^nx^n \right)
$$ could just be

$$
\frac{\ln(1+x)}{1+x},\quad |x|<1.
$$

A: In addition to the various comments about the overall sum, it's worth noting that there's a reasonable 'simplest form' for the coefficients as well.  Factor out the terms $x^{n+1}$ and note that you're looking at a sum of the form $a_{n+1}=(-1)^n\sum_{i=0}^n\frac1{i+1}$, where I've labeled this as $a_{n+1}$ since it's the coefficient of $x^{n+1}$ in the product series. By reindexing the sum and then making the substitution $n+1\mapsto n$, this is the same as $a_n=(-1)^{n-1}\sum_{i=1}^n\frac1i$, which should be a familiar-looking series.
