Closed Form formulas

Question

$\sum _{k=2}^{\infty} x^k/(k-1)k$

for $|x|<1$.

I'm supposed to find a closed form formula for this, but I don't know what closed form formula means. Is there a particular formula to follow? Thanks. :)

Answer 1

HINT

Let us call the sum

$$f(x)=\sum_{k=2}^{\infty} \frac{x^k}{k(k-1)}.$$

We see

$$f''(x)=\sum_{k=2}^{\infty} x^{k-2}= \frac{1}{1-x}$$ because we have a geometric series. Note that $|x|<1$ was given as a condition.

Hence to recover $f(x),$ we need to integrate $\frac{1}{1-x}$ twice.

As a result $$f(x)= \int_{0}^{x} \int_{0}^{y} \frac{1}{1-t} \ dt \ dy.$$ How can we evaluate the right hand side ? The right hand side is what is meant by a closed formula.

Remark We essentially are solving the differential equation $$f''(x)=\frac{1}{1-x}$$ with $f(0)=f'(0)=0.$ Such an equation is an Euler Cauchy equation, http://mathworld.wolfram.com/EulerDifferentialEquation.html.

Answer 2

$$\frac{1}{k(k-1)}=\frac{-1}{k}+\frac{1}{k-1}$$ $$\sum _{k=2}^{\infty} \frac{x^k}{(k-1)k}=-\sum _{k=2}^{\infty} \frac{x^k}{k}+\sum _{k=2}^{\infty} \frac{x^k}{k-1}$$ $\sum _{k=2}^{\infty} \frac{x^k}{k}=-x+\sum _{k=1}^{\infty} \frac{x^k}{k}=-x-\ln|1-x|$

$\sum _{k=2}^{\infty} \frac{x^k}{k-1}=x\sum _{k=1}^{\infty} \frac{x^k}{k}=-x\ln|1-x|$