Find the sum $\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)*(n+2)}$

Question

Need to find the sum $\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)(n+2)}$

---

What I did based on the suggestions:

1. Multiplied and divided by $x$
2. "forgot" the $x$ in the denominator and then took the derivative of the remaining expression 2 times: ended up getting $\frac{1}{x}\sum_{n=1}^{\infty} x^n$
3. Expanded this geometric series and then integrated the result once:

$\frac{1}{x}\sum_{n=1}^{\infty} x^n = \frac{1}{x}\int \frac{x}{1-x}dx = \frac{1}{x}(-x - \ln(1-x)) $

4. And then integrated again:

$\frac{1}{x}\int (-x - \ln(1-x))dx = \frac{1}{x}(-0.5x^2+x-x\ln(1-x)+\ln(1-x)) $ which is apparently a correct answer.

Is there any other more adequate way of solving this? Seems really non-intuitive to me, especially the double derivative-double integral part. Or maybe someone could shed the light regarding this concept ... like, why does it work in such a marvelous way

Answer 1

Note that the radius of convergence is $1$.

If $x=0$, the sum is $0$.

Otherwise, the Taylor series of $\ln(1-x)$ for $x \in [-1,1) \setminus \{0\}$ is$$\ln(1-x) = \sum_{n=0}^\infty\frac{x^{n+1}}{n+1} \tag{1}$$

$$\ln(1-x) = x\sum_{n=0}^\infty\frac{x^{n}}{n+1}=x\sum_{n=-1}^\infty \frac{x^{n+1}}{n+2}$$

$$\frac{\ln(1-x)}{x}-1 =\sum_{n=0}^\infty \frac{x^{n+1}}{n+2} \tag{2}$$

Use $(1)$ to subtract $(2)$,

$$\ln (1-x) - \frac{\ln (1-x)}{x}+1=\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)(n+2)}$$

$$\ln (1-x) - \frac{\ln (1-x)}{x}+1=\frac{x}{2}+\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)(n+2)}$$

$$\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)(n+2)}=-\ln (1-x) + \frac{\ln (1-x)}{x}-1-\frac{x}{2}$$

Answer 2

All the approaches I have read so far are essentially based around the fact that the radius of convergence of a (complex) power series $f(z)$ remains the same when integrating and differentiating $f(z)$ term by term. A possible approach (which is basically the same argument, written formally) is to find an ordinary differential equation that $f(z)$ satisfies and try to solve such equation on a suitable domain*.

That is, if you let $f(x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)(n+2)}$ then $f'(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n+2}$ for all $x\in (-1,1)$ by the first theorem I mentioned. Hence \begin{equation} x^2f'(x)=\sum_{n=1}^{\infty}\frac{x^{n+2}}{n+2}=\frac{1}{2} \left(-x^2-2 x-2 \log (1-x)\right) \end{equation} by the uniqueness of Taylor expansions. Therefore $f(x)$ satisfies the (trivial) ordinary differential equation $f'(x)=h(x)$ where $h(x)=\frac{1}{2x^2} \left(-x^2-2 x-2 \log (1-x)\right)$. Since $h\in C^{1}(0,1)$ one can apply the machinery from the theorems of existence and uniqueness for first-order ODE’s to find that \begin{equation} f(x)=-1-\frac{x}{2}-\log (1-x)+\frac{\log (1-x)}{x} \end{equation} in $(0,1)$.

---

*A theorem one could apply is the following. Let $\Omega\subset \mathbb{C}$ be a simply connected region and $z_0\in \Omega$. For any complex numbers $h_0,\dots,h_{n-1}$ there exists a unique holomorphic function $h$ on $\Omega$ such that \begin{equation} \frac{d^n h}{dz^n}+a_1\frac{d^{n-1} h}{dz^{n-1}}+\cdots+a_{n-1}\frac{dh}{dz}+a_nh=0 \end{equation} in $\Omega$ and moreover $h(z_0)=h_0$, $h'(z_0)=h_1$,$\dots$, $h^{(n-1)}(z_0)=h_{n-1}$.

Sadly, this theorem does not apply to the the non-homogenous ODE I that found.