I am trying to solve a math puzzle asked in a puzzle competition. The first part of the puzzle for me was to find an infinite sequence which is;
\begin{eqnarray*} \frac{7}{6^1}+\frac{28}{6^2}+\frac{84}{6^3}+\frac{210}{6^4}+\frac{462}{6^5}+.......... \end{eqnarray*}
The result of this summation will be the "expected value" of a series of events.
Since the sequence is getting smaller and smaller towards zero, it should have a valid sum.
The sequence is not an "arithmetic series" because it does not have "common difference".
It is not a "geometric series" because it does not have "common ratio".
It is not an "arithmetico geometric series" because it does not have any combination of common ratio and common difference.
The only relation I found between the elements of sequence is this;
\begin{eqnarray*} \frac{7}{6}+\left(\frac{7}{6}\right)\frac{8}{2}\frac{1}{6}+\left(\frac{7}{6}\frac{8}{2}\frac{1}{6}\right)\frac{9}{3}\frac{1}{6}+\left(\frac{7}{6}\frac{8}{2}\frac{1}{6}\frac{9}{3}\frac{1}{6}\right)\frac{10}{4}\frac{1}{6}+\left(\frac{7}{6}\frac{8}{2}\frac{1}{6}\frac{9}{3}\frac{1}{6}\frac{10}{4}\frac{1}{6}\right)\frac{11}{5}\frac{1}{6}+............. \end{eqnarray*}
So the notation of the relation between the elements of this infinite sequence is;
\begin{eqnarray*} \Pi_{j=0}^{k}\left(\frac{1}{6}\right)\left(\frac{j+7}{j+1}\right) \end{eqnarray*}
Then, the summation of all the elements is;
\begin{eqnarray*} \sum_{k=0}^{\infty}\Pi_{j=0}^{k}\left(\frac{1}{6}\right)\left(\frac{j+7}{j+1}\right) \end{eqnarray*}
Here I am done :)
I am not an expert mathematician, but a puzzle lover. I am not familiar with advanced mathematic and related notations. Because of the puzzle I am working on, I make deep dive to sequences and series concept. I don 't want the solution, I just need a clue to proceed.
Any help will be appreciated.. Regards..
Another way to write $$ \begin{align} \sum_{k=1}^\infty\prod_{j=0}^{k-1}\frac16\frac{j+7}{j+1} &=\sum_{k=1}^\infty\frac1{6^k}\binom{k+6}{k}\\ &=-1+\sum_{k=0}^\infty\left(-\frac16\right)^k\binom{-7}{k}\\ %&=\left(\frac65\right)^7-1\\[9pt] %&=\frac{201811}{78125} \end{align} $$ The last sum can be evaluated using the Generalized Binomial Theorem.