Taylor Series of $\frac{\log(z+1)}{z+1}, z_{0}=0 $ Finding the Taylor Series of
$\frac{\log(z+1)}{z+1}, z_{0}=0 $
Where Log is the complex logarithm.
At first, I tried to find the series for $\log(z+1)$ and $\frac{1}{z+1}$ and multiply them. However, I realized that it does not work because for series multiplication we cannot just add the exponents as if they are a single term.
Seeing the derivative of Log function,  $\frac{1}{z+1}$  in the denominator, I integrated the function and got $\frac{1}{2}\log(z+1)^{2}$. 
Does this mean I should find the Taylor series for $\log(z+1)$ first and square it?
Thanks.
 A: 
However, I realized that it does not work because for series multiplication we cannot just add the exponents as if they are a single term.

You should know how to do this properly, so try to compute:
$$\left[z-\frac{z^2}{2} + \frac{z^3}{3}-\frac{z^4}{4}+ \mathcal{O}(z^5)\right]\times\left[1-z + z^2-z^3+\mathcal{O}(z^4)\right]$$
to order $z^4$.
A: Hint: Let $f(z)=\frac{\log (z+1)}{z+1}$. Note that $f'(z)=\frac{1}{z+1}-f(z)$. Now can you find those coefficients you need in the Taylor Series?
A: Here is one way... I start with $x=-z$ to avoid confusion with signs.
$$
\log(1-x)=-\sum_{j=1}^\infty \frac{x^j}{j}
\\
\frac{1}{1-x}=\sum_{n=0}^\infty x^n
\\
\frac{\log(1-x)}{1-x} =
\big(-\sum_{j=1}^\infty \frac{x^j}{j}\big)\big(\sum_{n=0}^\infty x^n\big)
= -\sum_{k=1}^\infty\big(\sum_{j=1}^k \frac{x^j}{j}\;x^{k-j}\big)
\\
= \sum_{k=1}^\infty \big(-\sum_{j=1}^k\frac{1}{j}\big)x^k
=\sum_{k=1}^\infty (-H_k) x^k
$$
where $H_k$ is the "harmonic number"...  Finally,
$$
\frac{\log(1+z)}{1+z} = \sum_{k=1}^\infty (-1)^{k+1} H_k z^k
$$
