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Define for every $n \geq 2$ the function $$ f_n(x) = \frac{x^n}{n(n-1)}. $$ Aim: show that $\sum_{n=2}^\infty f_n(x) = x + (1-x)\log(1-x)$ for $|x|<1$.

Attempt: Firstly we find the ratio of convergence of this power series. Write $a_n = \frac{1}{n(n-1)}$. Then we have $$ \left| \frac{a_{n+1}}{a_n} \right| =\left|\frac{1}{(n+1)n} \cdot \frac{n(n-1)}{1} \right| = \left( \frac{n-1}{n+1} \right) \to 1 \textrm{ as } n \to \infty. $$ Hence we have have $R=1$.

Moreover, this convergence is uniform. Using the Weierstrass M-test, we have for $n \geq 2$ that $$ \left| \frac{x^n}{n(n-1)} \right| \leq \frac{1}{n(n-1)} =: M_n \textrm{ for all } x \in [0,1]. $$

The task now remains to show that the sum equals the expression given before. Let's first analyse the partial sums:
\begin{align*} s_n = \sum_{k=2}^n \frac{1}{k(k-1)} x^k = \sum_{k=2}^n \left( \frac{1}{k} - \frac{1}{k-1} \right) x^k = \sum_{k=2}^n \left( \frac{1}{k}x^k - \frac{1}{k-1}x^k \right) = \frac{1}{n}x^n - x^2. \end{align*} Hence, we have $$ \sum_{n=2}^\infty f_n(x) = \lim_{n \to \infty} s_n = -x^2 \lim_{n \to \infty} \frac{1}{n}x^n $$ Here I am stuck, it seems I'm not getting anywhere close to the desired output, and I'm not sure if what I did so far is correct at all. Any ideas appreciated, thanks.

Sigurd
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  • Showing if the series is convergent for the end points - totally pointless. Showing uniform convergence was pointless too. There's a theorem that goes as follows. You can interchange the order of summation and integration for the interior of the interval of convergence for power series. Use it twice, and integrate from $0$, to say, $y$ – Jakobian Jun 29 '18 at 18:57
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    @Adam: How can I know something is pointless if I haven't even solved the question yet?! At least I made an attempt. – Sigurd Jun 29 '18 at 18:59
  • I'm telling you now. It wasn't a personal assault, mind you. No need to get angry – Jakobian Jun 29 '18 at 19:00
  • Thanks for the help – Sigurd Jun 29 '18 at 19:00
  • Better give some constructive suggestions than saying that everything I wrote is 'totally' pointless – Sigurd Jun 29 '18 at 19:03

1 Answers1

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This seems overly complicated.

$f_n''=x^{n-2}$, so we can integrate $$(\sum_{n=2}^{\infty} f_n (x))''=\sum_{n=2}^{\infty} x^{n-2}=\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}$$

This holds for all $|x|<1$, as $1$ is the radius of convergence for $\sum f_n''$.

It only remains to integrate it twice to the result.

The way of life
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  • Thanks for the suggestion, I hadn't thought about this approach yet but that indeed makes it much easier. In this case, should we integrate from $2$ to $x$, since the sum starts at $2$? Or simply from $0$ to $x$? – Sigurd Jun 29 '18 at 19:15
  • $0$ to $x$ would be best, as the constants $f_n(0), f_n'(0)$ are $0$. – The way of life Jun 29 '18 at 19:18