Summation Formula for Series

Question

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.

Answer 1

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.

Answer 2

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 $q=r^{-1}:$

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

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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.