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I am interested in ways of evaluating the following infinite seris: $$ \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}. $$

I already know the answer from Wolfram Alpha but I would like to see some methods of evaluating it as I haven't been able to find many (any?) examples involving an infinite series with the $(m+k)!$ in the numerator and the $k!$ in the denominator, it seems that it is more common to find $k!$ in the numerator and $(m+k)!$ in the denominator.

So what are some methods that can be used to evaluate this series?

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This is the non-calculus way. Let the sum be $S_m$. Then, \begin{align} & 5S_m-S_m =\sum_{k=0}^{\infty}\frac{(m+k+1)\cdots(k+2)}{5^k}-\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+1)}{5^k}\\ =&(m+1)!+m\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+2)}{5^k}\\ =&(m+1)!+5m\sum_{k=2}^{\infty}\frac{(m-1+k)\cdots(k+1)}{5^k}\\ =& (m+1)!+5m\left(S_{m-1}-\frac{m!}{5}\right)=m!+5mS_{m-1}\\ \implies& S_m=\frac{m!}{4}+\frac{5m}{4}S_{m-1}\tag{1}\\ \implies&S_m=\frac{m!}{4}+\frac{5m}{4}\left(\frac{(m-1)!}{4}+\frac{5(m-1)}{4}S_{m-2}\right)\\ =&\frac{m!}{4}+\frac{5m!}{4^2}+\frac{5^2m(m-1)}{4^2}S_{m-2}\\ =&\frac{m!}{4}+\frac{5m!}{4^2}+\frac{5^2m!}{4^3}+\frac{5^3m(m-1)(m-2)}{4^3}S_{m-3}\\ =&\frac{m!}{4}\sum_{k=0}^n\left(\frac{5}{4}\right)^k+\frac{5^{n+1}m\cdots(m-n)}{4^{n+1}}S_{m-n-1} \end{align} Setting $n=m-1$ gives us, $$ S_m=\frac{m!}{4}\sum_{k=0}^{m-1}\left(\frac{5}{4}\right)^k+m!\left(\frac{5}{4}\right)^mS_0 $$ As $S_0$ is a geometric series with value $\frac{1}{4}$ our expression becomes, $$ S_m=\frac{m!}{4}\sum_{k=0}^{m}\left(\frac{5}{4}\right)^k=m!\left(\left(\frac{5}{4}\right)^{m+1}-1\right) $$

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Hint Consider the series $$x^m\sum_{k = 1}^\infty x^k = \sum_{k = 1}^\infty x^{m + k}$$ then do some differentiations. What you get?

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Here is another method and it is straightforward. Let $$ f(x)=\sum_{k=1}^\infty \frac{(m+k)!}{k!}x^k. $$ Then integrating $m$ gives \begin{eqnarray} F(x)&:=&\int \cdots\int f(x)dx\cdots dx\\ &=&\sum_{k=1}^\infty x^{m+k}+\sum_{k=0}^{m-1}C_kx^k\\ &=&\frac{x^{m+1}}{1-x}+\sum_{k=0}^{m-1}C_kx^k\\ &=&\frac1{1-x}\bigg[\sum_{k=1}^{m+1}\binom{m+1}{k}(x-1)^{m+1-k}+1\bigg]+\sum_{k=0}^{m-1}C_kx^k\\ &=&\frac1{1-x}-\sum_{k=1}^{m+1}\binom{m+1}{k}(x-1)^{m-k}+\sum_{k=0}^{m-1}C_kx^k\\ \end{eqnarray} where $C_0,C_1,\cdots,C_{m-1}$ are constants. Now differentiating $m$ times gives $$ f(x)=F^{(m)}(x)=\frac{m!}{(1-x)^{m+1}}-m!. $$ So $$ f(\frac15)=m!\bigg[\bigg(\frac54\bigg)^m-1\bigg]. $$

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  • $\begingroup$ There seems to be a mistake as differentiating the last line $m$ times doesn't work. This can be seen for the case of $m=1$. We get $$1/(1 - x)^2 + 1/(x - 1)^2$$ which is not equal to the answer given by the formula: $$1/(1 - x)^2 - 1$$ $\endgroup$ – csss Nov 9 '19 at 9:07
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    $\begingroup$ From 3rd to 4th line, $x^{m + 1}/(1 - x)$ can easlily be expanded to $$\frac{x^{m + 1}}{1 - x} = \frac{1}{1 - x} - \sum_{k = 1}^m x^k$$ by just writing $x^{m + 1}$ as $(x^{m + 1} - 1) + 1$. $\endgroup$ – Azlif Nov 9 '19 at 12:17
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    $\begingroup$ @csss Yes, the mistake is on the power of $x$ in the binomial expansion. They should be $k$ and $k - 1$ instead of $m + 1 - k$ and $m - k$, since $1$ is taken out of the sum ($k = m + 1$ will again give you $1$ instead of $(x - 1)^{m +1}$). Or the sum can also be changed to from $k = 0$ to $m$ instead starting from $1$ to $m+1$. $\endgroup$ – Azlif Nov 9 '19 at 12:35
  • $\begingroup$ @Azlif Ok thanks, by the way, I think the sum should start from $0$ instead of $1$ in your comment? Also, do you know how the binomial expansion arises in line 4? $\endgroup$ – csss Nov 9 '19 at 15:07
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    $\begingroup$ @csss Ah yes you're right, it should start from $0$. xpaul wrote $x^{m + 1}$ as $[(x - 1) + 1]^{m+1}$ and then he used binomial expansion. $\endgroup$ – Azlif Nov 9 '19 at 15:18

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