Need help to prove the summation of series. It works in programming, but I don't know how to prove.

Question

It produces the series

$$ \begin{cases} a_0 &=\frac{k}{m}\\ a_x &= a_{x-1} \cdot \frac{m+x-k}{m+x}, x \in [1, 2 \dots] \end{cases} $$

It seems that (how to prove?)

$$\sum_{i=0}^{\infty} a_i = \frac{k}{k-1}$$

I have tested it using the python code, it works

import numpy as np
# Change the k and m, and make sure m > k
k = 7
m = 10
array = [k/m]
for x in range(1, 1000):
    v = array[x-1] * (m+x-k) / (m+x)
    array.append(v)
print(array)
It produces two values which are the same.
print(np.sum(array), k/(k-1))
```

Answer 1

You can use telescoping series: \begin{align*} \sum_{x=0}^\infty a_x &= \frac{k}m \sum_{x=0}^\infty \frac{(m+x-k)!/(m-k)!}{(m+x)!/m!}\\ &= \frac{k}m \frac{m!}{(m-k)!}\sum_{x=0}^\infty \frac1{(m+x)...(m+x-k+1)}\\ &= \frac{k}m \frac{m!}{(m-k)!}\frac1{k-1}\sum_{x=0}^\infty \frac{(m+x)-(m+x-k+1)}{(m+x)...(m+x-k+1)}\\ &= \frac{k}m \frac{m!}{(m-k)!}\frac1{k-1}\sum_{x=0}^\infty \left[\frac1{(m+x-1)...(m+x-k+1)}-\frac1{(m+x)...(m+x-k+2)}\right] \\ &= \frac{k}m \frac{m!}{(m-k)!}\frac1{k-1}\left[\sum_{x=0}^\infty \frac1{(m+x-1)...(m+x-k+1)}-\sum_{x=1}^\infty\frac1{(m+x-1)...(m+x-k+1)}\right] & (*)\\ &= \frac{k}m \frac{m!}{(m-k)!}\frac1{k-1}\frac1{(m-1)...(m-k+1)}\\ &= \frac{k}m \frac{m!}{(m-k)!}\frac1{k-1}\frac{(m-k)!}{(m-1)!}\\ &=\frac{k}{k-1} \end{align*} where in (*), all the terms except for the left $x=0$ term cancel.

Answer 2

This is a special case of Gauss' hypergeometric summation $$\sum_{x=0}^\infty\frac{(a)_x(b)_x}{x!(c)_x}=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},$$ where $$(a)_x=a(a+1)\dotsm(a+x-1)$$ and $\mathrm{Re}(c-a-b)>0$. Indeed, you have $$a_x=\frac km\frac{(m+1-k)_x}{(m+1)_x},$$ so your series is $$\frac km\sum_{x=0}^\infty\frac{(1)_x(m+1-k)_x}{x!(m+1)_x}=\frac km\frac{\Gamma(m+1)\Gamma(k-1)}{\Gamma(m)\Gamma(k)}=\frac k{k-1},$$ where $\mathrm{Re}(k)>1$.

Answer 3

The $a_x, x\geq 0$ can also be represented by inverse binomial coefficients. We then use the integral representation \begin{align*} \binom{p}{q}^{-1}=(p+1)\int_{0}^1 t^q(1-t)^{p-q}\,dt\tag{1} \end{align*} to simplify the series and derive: \begin{align*} \color{blue}{\sum_{x=0}^{\infty}a_x=\frac{k}{k+1}}\tag{2} \end{align*} We obtain \begin{align*} {\color{blue}{a_x}}&=a_{x-1}\,\frac{m+x-k}{m+x}=a_{x-2}\,\frac{m+x-1-k}{m+x-1}\cdot\frac{m+x-k}{m+k}\\ &=a_0\frac{m+x-(x-1)-k}{m+x-(x-1)}\cdots\frac{m+x-1-k}{m+x-1}\cdot\frac{m+x-k}{m+k}\\ &=a_0\prod_{q=0}^{x-1}\frac{m+x-q-k}{m+x-q}=a_0\frac{(m+x-k)!}{(m-k)!}\cdot\frac{m!}{(m+x)!}\\ &\,\,\color{blue}{=\frac{k}{m}\binom{m}{k}\binom{m+x}{k}^{-1}}\tag{3} \end{align*} We use (1) and (3) and transform the series as follows: \begin{align*} {\color{blue}{\sum_{x=0}^{\infty}a_x}}&=\frac{k}{m}\binom{m}{k}\sum_{x=0}^{\infty}\binom{m+x}{k}^{-1}\\ &=\frac{k}{m}\binom{m}{k}\sum_{x=0}^{\infty}(m+x+1)\int_{0}^{1}t^k(1-t)^{m+x-k}\,dt\\ &=\frac{k}{m}\binom{m}{k}\int_{0}^1\left(\frac{t}{1-t}\right)^k \sum_{x=0}^{\infty}(m+x+1)(1-t)^{m+x}\,dt\tag{4}\\ \end{align*} Since \begin{align*} {\color{blue}{\sum_{x=0}^{\infty}(m+x+t)(1-z)^{m+x}}}&=-\frac{d}{dt}\sum_{x=0}^{\infty}(1-t)^{m+x+1}\\ &=-\frac{d}{dt}(1-t)^{m+1}\sum_{x=0}^{\infty}(1-t)^x\\ &=-\frac{d}{dt}(1-t)^{m+1}\frac{1}{1-(1-t)}\\ &=-\frac{d}{dt}\,\frac{(1-t)^{m+1}}{t}\\ &\color{blue}{=\frac{1+mt}{t^2}(1-t)^m} \end{align*} we can put it into (4) and get \begin{align*} {\color{blue}{\sum_{x=0}^{\infty}a_x}}&=\frac{k}{m}\binom{m}{k}\int_{0}^1\left(\frac{t}{1-t}\right)^k \frac{1+mt}{t^2}(1-t)^m\,dt\\ &=\frac{k}{m}\binom{m}{k}\int_{0}^1(1+mt)(1-t)^{m-k}t^{k-2}\,dt\\ &=\frac{k}{m}\binom{m}{k}\left(\int_{0}^1(1-t)^{m-k}t^{k-2}\,dt +m\int_{0}^1(1-t)^{m-k}t^{k-1}\,dt\right)\\ &=\frac{k}{m}\binom{m}{k}\left(\frac{1}{m-1}\binom{m-2}{k-2}^{-1}+\binom{m-1}{k-1}^{-1}\right)\\ &=\frac{k}{m}\binom{m}{k}\left(\frac{1}{m-1}\cdot\frac{(k-2)!(m-k)!}{(m-2)!}+\frac{(k-1)!(m-k)!}{(m-1)!}\right)\\ &=\frac{k}{m}\binom{m}{k}\frac{(k-2)!(m-k)!}{(m-1)!}\left(1+(k-1)\right)\\ &=\frac{k}{m}\binom{m}{k}\frac{m}{k-1}\binom{m}{k}^{-1}\\ &\,\,\color{blue}{=\frac{k}{k-1}} \end{align*} and the claim (2) follows.

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Note: Another way to use the inverse binomial coefficient (1) is to recall the recursion formula \begin{align*} \color{blue}{\binom{p}{q}^{-1}-\binom{p+1}{q}^{-1}=\frac{q}{q+1}\binom{p+1}{q+1}^{-1}} \end{align*} and use with (3) the telescopic property in \begin{align*} \sum_{x=0}^{\infty}a_x&=\frac{k}{m}\binom{m}{k}\sum_{x=0}^{\infty}\binom{m+x}{k}^{-1}\\ &=\frac{k}{m}\binom{m}{k}\sum_{x=0}^{\infty} \left(\binom{m+x-1}{k-1}^{-1}-\binom{m+x}{k-1}^{-1}\right)\frac{k}{k-1} \end{align*} and proceed in a similar way to the top-scoring answer.

Answer 4

Supposing we are interested in

$$\sum_{q\ge 0} {m+q\choose k}^{-1}.$$

Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$

$$\frac{1}{k} {n\choose k}^{-1} = [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$

We can re-write this as

$${n-1\choose k-1}^{-1} = n [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$

We get for our sum writing ${m+q\choose m+q-k}$

$$\sum_{q\ge 0} (m+q+1) [v^{m+q+1}] \log\frac{1}{1-v} (v-1)^k \\ = \sum_{q\ge m+1} q [v^q] \log\frac{1}{1-v} (v-1)^k = \;\underset{v}{\mathrm{res}}\; \log\frac{1}{1-v} (v-1)^k \sum_{q\ge m+1} \frac{q}{v^{q+1}} \\ = - \;\underset{v}{\mathrm{res}}\; \log\frac{1}{1-v} (v-1)^k \left(\sum_{q\ge m+1} \frac{1}{v^q}\right)' \\ = - \;\underset{v}{\mathrm{res}}\; \log\frac{1}{1-v} (v-1)^k \left(\frac{1}{v^{m+1}} \frac{1}{1-1/v}\right)'.$$

We find for the derivative

$$\left(\frac{1}{v^{m}} \frac{1}{v-1}\right)'.$$

This produces two pieces, the first is

$$\;\underset{v}{\mathrm{res}}\; \log\frac{1}{1-v} (v-1)^{k-1} \frac{m}{v^{m+1}} \\ = m [v^m] \log\frac{1}{1-v} (v-1)^{k-1} = {m-1\choose m-k}^{-1} = {m-1\choose k-1}^{-1}.$$

and the second

$$\;\underset{v}{\mathrm{res}}\; \log\frac{1}{1-v} (v-1)^{k-2} \frac{1}{v^m} \\ = [v^{m-1}] \log\frac{1}{1-v} (v-1)^{k-2} = \frac{1}{m-1} (m-1) [v^{m-1}] \log\frac{1}{1-v} (v-1)^{k-2} \\ = \frac{1}{m-1} {m-2\choose m-k}^{-1} = \frac{1}{m-1} {m-2\choose k-2}^{-1}.$$

Adding the two contributions

$${m\choose k}^{-1} \left[ \frac{m}{k} + \frac{1}{m-1} \frac{m(m-1)}{k(k-1)} \right] = {m\choose k}^{-1} \frac{m}{k} \left[1+ \frac{1}{k-1} \right] = {m\choose k}^{-1} \frac{m}{k-1}.$$

This confirms the result by @MarkusScheuer.

Answer 5

Hint: prove that $$ \sum_{i=0}^y a_i = \frac k{k-1} \biggl( 1 - \prod_{j=0}^y \biggl( 1 - \frac{k-1}{m+j} \biggr) \biggr). $$