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I have a series of the form : \begin{equation} \frac{1}{M-1} + \frac{q}{M-2} + \frac{q^2}{M-3} + \frac{q^3}{M-4} + \frac{q^4}{M-5}+\dots = \sum_{i=1} ^{M-1} \frac{q^{i-1}}{M-i} \end{equation} I want to solve this series to find a general formula that provides its sum. I am not able to figure out the best and easy way to proceed with this. I would be glad if anybody could point me the right direction for solving such series.

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  • $\begingroup$ Please, edit your question. It's not clear. $\endgroup$ – user399481 Dec 21 '16 at 17:20
  • $\begingroup$ what happen with $M$ and $q$ are they bounded? $\endgroup$ – Camilo Acevedo. Dec 21 '16 at 17:24
  • $\begingroup$ @AashishDhungana I've edited your problem could you double check I captured it properly? $\endgroup$ – Kitter Catter Dec 21 '16 at 17:27
  • $\begingroup$ @CamiloAcevedo q is a probability variable between [0-1] and M in my case is 100. I want to generalize the series for any M. Does this help ? $\endgroup$ – Aashish Dhungana Dec 21 '16 at 17:31
  • $\begingroup$ @Kitter: yes, thats is correct and i ranges from 1 to M-1. $\endgroup$ – Aashish Dhungana Dec 21 '16 at 17:36
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Your series may be written as $$ \sum_{n=0}^\infty \frac{q^n}{M-n-1},\quad |q|<1, \,M \neq 1,2,\cdots, $$ this is an instance of the Lerch transcendent function $\Phi$: $\Phi(q,1,1-M)$. For general parameters $q$ and $M$, there is no known simple closed form of it.

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    $\begingroup$ I am sorry for not providing the full information about the problem. It is not actually an infinite series. I have edited the question now. $\endgroup$ – Aashish Dhungana Dec 21 '16 at 17:53
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Sum it from $M-1$ to $1$, i.e., sum it all up backwards.

$$\sum_{k=1}^{M-1}\frac{q^{i-1}}{M-i}=\sum_{k=1}^{M-1}\frac{q^{M-k-1}}{k}=q^{M-1}\sum_{k=1}^{M-1}\frac{q^{-k}}{k}$$

Let $q=r^{-1}:$

$$=r^{1-M}\sum_{k=1}^{M-1}\frac{r^k}{k}\tag{$\star$}$$


Recall the geometric series:

$$\frac{1-x^{M-1}}{1-x}=\sum_{k=1}^{M-1}x^{k-1}$$

Integrate wrt $x$ from $0$ to $r:$

$$\int_0^r\frac{1-x^{M-1}}{1-x}\ dx=\int_0^r\sum_{k=1}^{M-1}x^{k-1}\ dx=\sum_{k=1}^{M-1}\frac{r^k}k$$

Thus, you may rewrite your sum as

$$\sum_{k=1}^{M-1}\frac{q^{i-1}}{M-i}=q^{M-1}\int_0^{1/q}\frac{1-x^{M-1}}{1-x}\ dx$$

From there, you may use integration techniques to derive closed form solutions for some $q,M$.

As demonstrated on this graph.

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