Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.
I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.
 A: \begin{align*}
\frac{\ln(1+x)}{1+x} 
    &= \ln(1- -x) \cdot \frac{1}{1- -x}  \\
    &= \left( \sum_{i=1}^\infty \frac{-(-x)^i}{i} \right) \left( \sum_{j=0}^\infty (-x)^j \right),  & |x| < 1\\
    &=_{\text{why?}} \sum_{i=1}^\infty \sum_{j=0}^\infty \frac{-(-x)^i}{i} (-x)^j  \\
    &= \sum_{i=1}^\infty \sum_{j=0}^\infty \frac{(-1)^{i+j+1}}{i} x^{i+j}
\end{align*}
Now change variables to $k = i+j$ and $\ell$.  On a diagram of the terms in this double series (a plot of points on the $(i,j)$-plane corresponding to terms in the series), the lines of constant $k$ have slope $-1$ and meet $k$ points, so we may take $\ell = i$, running from $1$ to $k$.  (Again, this is vastly easier to see in the diagram.)  Now make $k$ be the outer sum and $\ell$ be the inner sum and you will immediately have your result,
$$  \sum_{k=1}^\infty \left( -(-1)^k x^k \sum_{\ell =1}^k \frac{1}{\ell} \right)  \text{.}  $$
A: Multiply the two series $\ln(1+x) = x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \cdots$ and $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$.
For instance, the coefficient of $x^2$ in the product is $(-\frac{1}{2} x^2) \cdot 1 + x \cdot (-x) = -(1 - \frac{1}{2}) x^2$.
A: You must multiply the series
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac 1{1+x}=1-x+x^2-x^3+x^4-x^5+O\left(x^6\right)$$
$$\frac{\log(1+x)}{1+x}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) \right) \times$$ $$\left(1-x+x^2-x^3+x^4-x^5+O\left(x^6\right) \right) $$
$$\frac{\log(1+x)}{1+x}=x-\frac{3 x^2}{2}+\frac{11 x^3}{6}-\frac{25 x^4}{12}+\frac{137
   x^5}{60}+O\left(x^6\right)$$
In fact, almost as you wrote in the post
$$\frac{\log(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n+1} H_n \,x^n$$
A: All the other answers to your question focus on how to compute the Maclaurin series of this function using the known series. I will sketch a strategy to compute it by directly computing Maclaurin coefficients. Let $$(1+x)y=\ln(1+x)$$ for all $x\gt -1. $ Then lets look at higher derivatives to see a possible pattern: $$(1+x)\dfrac{dy}{dx}+y=\dfrac{1}{1+x}\tag1$$
$$(1+x)\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}=-\dfrac{1}{(1+x)^2}\tag2$$
$$(1+x)\dfrac{d^3y}{dx^3}+3\dfrac{d^2y}{d^2x}=\dfrac{2}{(1+x)^3}\tag3.$$ By looking at these, it is not difficult to guess that
$$(1+x)\dfrac{d^{n+1}y}{dx^{n+1}}+(n+1)\dfrac{d^ny}{d^nx}=\dfrac{(-1)^{n}n!}{(1+x)^{n+1}}\qquad \forall n\in\mathbb{N}\cup\{0\}.$$ We can use mathematical induction to establishes our claim rigorously. Since coefficients of Maclaurin series are given by
$$a_n=\dfrac{1}{n!}\dfrac{d^ny}{d^nx}\Big\vert_{x=0},$$ we can show that $a_n=(-1)^{n+1}H_n$ by solving the recursive relation $$a_{n+1}+a_n=\dfrac{(-1)^n}{n+1},\qquad a_0=0.$$ (You can easily solve this recursion by multiplying whole equation by $(-1)^n$ and then telescoping). In fact, we can easily use it to compute Taylor coefficients at any point we want.
