Krdytkyu

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-xln(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

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$$ln(1-x) = sum_{n=0}^inftyfrac{x^{n+1}}{n+1} tag{1}$$

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

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

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

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

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

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

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 $xin (-1,1)$ 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 $hin C^{1}(0,1)$ 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 $Omegasubset mathbb{C}$ be a simply connected region and $z_0in Omega$. 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_1frac{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.