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Mathematica tells me that $$ \sum_{n=1}^{\infty}n\prod_{k=1}^{n}\frac{1}{1+ka} = \frac{1}{a} $$ I could prove it for $a\rightarrow 0$, $a=1$ and $a\rightarrow \infty$, but could not find a general proof for real $a>0$. Can anyone find one? Many thanks!

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Note, you night as well write it as $\prod_{k=1}^n$, since $k=0$ adds nothing. – Thomas Andrews Jul 8 '14 at 20:46
Sure, I edited accordingly – citronrose Jul 8 '14 at 21:11
up vote 10 down vote accepted

Lemma: \begin{align} \sum_{n=1}^{N} \frac{n}{\prod_{k=1}^n(1+ka)}=\frac{1}{a}\left[1-\frac{1}{\prod_{k=1}^N(1+ka)}\right]. \end{align} Proof: Induction on $N$.

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(+1) Just after an almost magical proof through integration by parts, this comes as a shot in the heart to me :D – Jack D'Aurizio Jul 8 '14 at 23:18
@JackD'Aurizio I initially reduced this to a hypergeometric series but then noticed a very clear pattern for the finite sum. – Start wearing purple Jul 8 '14 at 23:20
+1 Alternatively see it as a telescoping sum: $$\frac{na}{\prod_1^n (1+ka)} = \frac{1}{\prod_{1}^{n-1} (1+ka)}-\frac{1}{\prod_{1}^{n} (1+ka)}$$ – Thomas Andrews Jul 8 '14 at 23:29
Wow, very elegant! Thank you so much! – citronrose Jul 9 '14 at 8:35
@citronrose Please feel free to use without any acknowledgement! – Start wearing purple Jul 16 '14 at 20:15

Simply a heads-and-tails game in disguise...

Consider a heads-and-tails game such that the probability of success of the $n$th throw is $$p_n=\frac{na}{1+na}.$$ Then the probability that the $n$th throw is the first success is $$ p_n\prod_{k=1}^{n-1}(1-p_k)=na\prod_{k=1}^n\frac1{1+ka}, $$ hence $$ \sum_{n=1}^{\infty}n\prod_{k=1}^{n}\frac{1}{1+ka}=\frac1a\,P(A), $$ where $A$ is the event that at least one success happens, ever. The event that no success amongst the $n$ first throws has probability $$ \prod_{k=1}^n(1-p_k)=\prod_{k=1}^n\frac1{1+ka}, $$ which converges to $0$ hence $P(A)=1$, QED.

Nota: This interpretation also yields the identity $$\sum_{n=1}^Nn\prod_{k=1}^{n}\frac{1}{1+ka}=\frac1a\,\left(1-\prod_{n=1}^{N}\frac{1}{1+na}\right), $$ since $a$ times the LHS is the probability that the first success is seen amongst the $N$ first throws, and the probability of the complementary event is, by independence, the product in the RHS.

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Very nice interpretation, it actually provides a better intuitive understanding of my problem. – citronrose Jul 9 '14 at 9:01
Replacing $na$ by $a_{n}$ we get the more general results that $$ \sum_{1}^{\infty} a_{n} \prod_{k=1}^{n} \frac{1}{1+a_{k}} = 1$$ (and $$\sum_{1}^{\infty} a_{n} \prod_{k=1}^{n} \frac{1}{1+a_{k}} = 1-\prod_{n=1}^{N}\frac{1}{1+a_{n}}$$) for any sequence $a_{n}>0$, right? – citronrose Jul 11 '14 at 8:26
The first identity is valid if and only if the series $\sum\limits_na_n$ diverges. The sum in the LHS of the second identity should end at $N$, not at $\infty$. Then yes, everything works as in my answer. – Did Jul 11 '14 at 14:24

We have that: $$A_n=\prod_{k=0}^n\frac{1}{1+ka} =\frac{1}{n!}\prod_{k=1}^{n}\frac{1}{a+\frac{1}{k}},$$ that can be regarded as a meromorphic function of the variable $a$. Moreover, for any $j\in[1,n]$: $$\begin{eqnarray*}\operatorname{Res}\left(A_n,a=-\frac{1}{j}\right)&=&\frac{1}{n!}\prod_{\substack{k=1\\k\neq j}}^{n}\frac{1}{-\frac{1}{j}+\frac{1}{k}}=(-1)^n j^{n-2}\prod_{\substack{k=1\\k\neq j}}^{n}\frac{1}{k-j}\\&=&\frac{(-1)^n\,j^{n-2}}{(-1)^{j-1}(j-1)!(n-j)!}\\&=&-\frac{(-1)^{n-j}\, j^{n-1}}{n!}\binom{n}{j}\end{eqnarray*} $$ hence the residue of $\sum_{n=1}^{+\infty}n A_n$ in $a=-\frac{1}{j}$ is: $$\begin{eqnarray*}-\sum_{n\geq j}\frac{(-1)^{n-j}\, j^{n-1}}{(n-1)!}\binom{n}{j}&=&-j!\cdot\frac{d^j}{dx^j}\left.\sum_{n\geq j}\frac{(-1)^{n-j}\, j^{n-1} x^n}{(n-1)!}\right|_{x=1}\\&=&-j!\cdot\frac{d^j}{dx^j}\left.\sum_{n=0}^{+\infty}\frac{(-1)^{n-j}\, j^{n-1} x^n}{(n-1)!}\right|_{x=1}\\&=&j!\cdot(-1)^j\frac{d^j}{dx^j}\left.\left(x\,e^{-jx}\right)\right|_{x=1}=0,\end{eqnarray*}$$ and the calculations you made for the cases $a=1$ and $a\to 0$ are enough to state $$\sum_{n=1}^{+\infty} n A_n = \frac{1}{a}$$ for any $a>0$. In another fashion, using the Euler Beta function: $$\begin{eqnarray*}\sum_{n=1}^{+\infty}n A_n &=& \sum_{n\geq 1}\frac{n}{a^n}\cdot\frac{\Gamma(1+1/a)}{\Gamma(1+1/a+n)}=\int_{0}^{1}(1-x)^{1/a}\sum_{n\geq 1}\frac{n x^{n-1}}{a^n \Gamma(n)}dx\\&=&\int_{0}^{1}(1-x)^{1/a}e^{x/a}\frac{x+a}{a^2}dx\\&=&\int_{0}^{1}(1-x)^{1/a}\frac{d}{dx}\left(\frac{x}{a}e^{x/a}\right)dx\\&=&\frac{1}{a}\int_{0}^{1}(1-x)^{1/a-1}\frac{x}{a}e^{x/a}dx\\&=&\frac{1}{a}\int_{0}^{1}(1-x^a)\exp\left(\frac{1-x^a}{a}\right)dx=\frac{1}{a},\end{eqnarray*}$$ since we have a cancellation induced by integration by parts: $$\begin{eqnarray*}\int_{0}^{1}x^a \exp\left(\frac{1-x^a}{a}\right)dx&=&\int_{0}^{1}x\cdot\left(x^{a-1}\exp\left(\frac{1-x^a}{a}\right)\right)dx\\&=&\left.-x\exp\left(\frac{1-x^a}{a}\right)\right|_{0}^{1}+\int_{0}^{1} \exp\left(\frac{1-x^a}{a}\right)dx.\end{eqnarray*}$$

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