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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 $f(1)$?

2 Answers 2

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Try using Abel's theorem.

  • 0
    @dawson: it's also a result of Abel's theorem since (atleast in my book) it is shown in Abel's theorem proof that if a power series converges at $x=\pm R$ then it is uniformly convergent in $[-R,R]$.2011-01-24
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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 $x = -1$.]