# Finding the sum $f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$

I'm trying to find $$f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$$

I found the radius of convergence of the above series which is $R=1$. Checking $x=\pm 1$ also yields a convergent series. Therefore the above series converges for all $x\in [-1, 1]$.

Using differentiation of the series term by term we get: $$f'(x)=\sum_{n=2}^{\infty} \frac{x^{n-1}}{n-1}=\sum_{n=1}^{\infty} \frac{x^{n}}{n}=-\log(1-x)$$ which also has $R=1$, and then, by integrating term by term we get: $$f(x)=\int_{0}^{x} f'(t)dt=-\int_{0}^{x} \log(1-t)dt=x-(x-1)\ln(1-x)$$

if I understand the theorems in my textbook correctly, the above formulas are true only for $x \in (-1, 1)$. It seems the above is correct since this is also what WolframAlpha says.

I'm abit confused though. At first, it seemed the above series converges for all $x\in [-1, 1]$ but in the end I only got $f(x)$ for all $|x|\lt 1$, something seems to be missing. What can I say about $f(-1)$ and $f(1)$?

-

Try using Abel's theorem.

-
I thought of Abel's theorem but I'm not sure on its usage, can I say that: $$f(1)=\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}(x-(x-1)\ln(1-x))=1$$ and $$f(-1)=\lim_{x\to -1^{+}}f(x)=\lim_{x\to 1^{+}}(x-(x-1)\ln(1-x))=\ln4-1$$ –  dawson Jan 22 '11 at 17:55
If so, why do I need Tauber's theorem (it's not in my textbook)? –  dawson Jan 22 '11 at 17:56
@dawson: No you don't need it. I am just having a braindead moment. –  Aryabhata Jan 22 '11 at 18:00
@dawson: Since you noted the series is convergent, Abel's theorem tells you that the lim of f is same as the limit of the series. So you seem to have it right :-) –  Aryabhata Jan 22 '11 at 18:07
On second thought, $\frac{|x|^n}{n(n-1)}\leq\frac{1}{n(n-1)}$ and $\sum \frac{1}{n(n-1)}$ converges, then by Weierstrass test the above series converges uniformly in $[-1, 1]$, does that make sense? –  dawson Jan 22 '11 at 18:22
show 3 more comments

If you rewrite $\frac{1}{n(n-1)}$ in the form $\frac{1}{n-1}-\frac{1}{n}$, then you can rewrite the series in both cases $x = \pm 1$ and compute their values directly. You can then confirm that in both cases the value you compute coincides with the value $f(\pm 1)$. (In other words, rather than appealing to Abel's theorem, as Moron suggests, in this particular case you can verify it.)

[Caveat: In the case $x = -1$, you will need to use the familiar series for $\log 2$, and maybe the easiest way to prove this is by appealing to Abel's theorem (applied to the series for $\log (1 + x)$). So my approach probably doesn't really avoid Abel's theorem, at least for $x = -1$.]

-