1

Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.

This is my idea.

We have the radius of convergence is $R=1$. And $\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.

Then let $f(x)=\sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.

We have $f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ and $f''(x)=\sum_{n=1}^{\infty} x^{n-1}=\dfrac{1-x^n}{1-x}$

Then if I take the primity of $f''(x)$, I get: $f'(x)=\displaystyle\int\limits_{0}^{x}\dfrac{1-t^n}{1-t}\mathrm{d}t=-\ln(1-x)-\displaystyle\int\limits_{0}^{x}\dfrac{t^n}{1-t}\mathrm{d}t$

Then I will get a sum again, and that's not the goal.

What can I do then? Thank you so much.

Did
  • 279,727
anvo
  • 95
  • 1
    The remaining integral in the final line vanishes when you take the limit, as the integrand vanishes. – Jack Tiger Lam Oct 23 '16 at 03:58
  • 2
    Your $f''(x)$ should be $1/(1-x)$, not $(1-x^n)/(1-x)$. (In particular, it should be a function of only $x$, not $x$ and $n$. The latter function is one of the partial sums.) – Ravi Fernando Oct 23 '16 at 04:00

1 Answers1

4

If you know (or can assume) the logarithm series for $|x|<1$:

$$-\log(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$$

then you can forego the derivation/integration steps by noticing that:

$$ \begin{align} \sum_{n=1}^{\infty} \frac{x^{n}}{n(n+1)} & = \sum_{n=1}^{\infty} x^n \left(\frac{1}{n} - \frac{1}{n+1} \right) \\ & = \sum_{n=1}^{\infty} \frac{x^n}{n} - \frac{1}{x} \sum_{n=2}^{\infty} \frac{x^n}{n} \\ & = -\log(1-x) - \frac{1}{x}(-\log(1-x) - x) \\ & = \left(\frac{1}{x}-1\right)\log(1-x) \;+\; 1 \end{align} $$

dxiv
  • 76,497