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$a_1,a_2,a_3,...$ is a sequence of real numbers such that for $n\in \mathbb{N}$: $$\sum_{k=0}^{n-1}\frac{a_{n-k}}{k+1}=\frac{1}{n+1}$$ How can I prove: $$a_n=\int_1^\infty{\frac{dx}{x^n(\pi^2+\ln^2(x-1))}}$$

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Can you tell the source of this problem. Also, have you tried induction? – dexter04 Feb 10 '13 at 5:12
in fact the solution must not be given. the equence must be derived. However the integral when replaced must be equal to $\frac{1}{n+1}$. But it may need complex analysis usage. – user59671 Feb 10 '13 at 5:16
The recursive definition of the $a_n$ is equivalent to the generating function identity $$\bigg( \sum_{m=0}^\infty \frac{y^m}{m+1} \bigg) \bigg( \sum_{m=1}^\infty a_my^m \bigg) = \sum_{m=1}^\infty \frac{y^m}{m+1},$$ which gives $$\sum_{m=1}^\infty a_my^m = 1 + \frac y{\ln(1-y)}.$$ I couldn't figure out how this helps though. – Greg Martin Feb 10 '13 at 7:12
Note that if $a_0 = -1$, then for each $n \geq 1$, $\frac{a_n}{1} + \frac{a_{n - 1}}{2} + \dotsb + \frac{a_1}{n} + \frac{a_0}{n + 1} = 0$. Hope this helps... – Herng Yi Feb 10 '13 at 7:59
@GregMartin, the series for $\ln(1 - y) / y$ is known, from it you can get it's reciprocal, and you have OP's answer. Probably in terms of a recurrence, though. – vonbrand Feb 10 '13 at 17:51
up vote 5 down vote accepted

As already noted by Greg Martin, if you set $$ \phi(z):=\sum_{n\ge 1} a_n z^n, $$ then the recurrence immediately gives $$ \phi(z) (-\frac{\log (1-z)}{z})=-\frac{\log (1-z)}{z}-1. $$ Therefore, $$ \phi(z)=1+\frac{z}{\log (1-z)}, $$ so, for $n\ge 1$, $$ a_n=[z^n] \frac{z}{\log (1-z)}. $$ This can be rewritten as $$ a_n=\frac{1}{2\pi i} \int \frac{1}{z^{n} \log (1-z)} \, dz \qquad (1) $$ where the integral is taken over a small circle $C$ around the origin, traversed in the counterclockwise direction.

Let the function $z\mapsto \log (1-z)$ have a branch cut along the positive real axis starting at $z=1$. Let $C'$ be the integration path $L\cup D\cup L'\cup E$, where $L$ is the line segment from $1+\epsilon+\epsilon i$ to $M+\epsilon i$ (on one side of the branch cut), $D$ is the arc of the circle $|z|=\sqrt{M^2+\epsilon^2}$ from $M+\epsilon i$ to $M-\epsilon i$, $L'$ is the line segment from $M-\epsilon i$ to $1+\epsilon-\epsilon i$ (on the other side of the branch cut), and $E$ is the arc of the circle $|z-1|=\sqrt{2}\epsilon $ from $1+\epsilon-\epsilon i$ to $1+\epsilon+\epsilon i$. The path is traversed in an overall counterclockwise direction around the origin and never crosses the branch cut.

Since $C$ can be deformed into $C'$ within the region where the integrand is analytic, the integration path in $(1)$ can be changed from $C$ to $C'$ without affecting the value of the integral. Let $\epsilon\to 0$. Then the integral on $E$ is $O(\epsilon |\log \epsilon|^{-1})$ and so approaches $0$, the integral over $L$ approaches $$ \int_1^M \frac{1}{z^n (\log(z-1)-\pi i)} \, dz,\qquad (2) $$ and the integral over $L'$ approaches $$ -\int_1^M \frac{1}{z^n (\log(z-1) + \pi i)} \, dz.\qquad (3) $$ Since the integral on $D$ is $O(M^{-(n-1)} (\log M)^{-1})$, combining $(1)$, $(2)$ and $(3)$ gives $$a_n=\int_1^M \frac{1}{z^n (\log^2 (z-1) + \pi^2)} \, dz+O(M^{-(n-1)} (\log M)^{-1}).$$ Now let $M\to\infty$ to give the desired result.

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