Evaluating Factorial Series by Rewriting the Numerator in the Factorial's Argument

Intuition

A series whose general term is $\dfrac{P(n)}{(an+b)!}$, with $P$ a polynomial in $n$, can almost always be summed in closed form. The trick is that a factorial cancels its own argument: $\dfrac{an+b}{(an+b)!}=\dfrac{1}{(an+b-1)!}$. So if you rewrite the numerator $P(n)$ as a combination of the falling factors $(an+b),(an+b-1),\dots$ of the denominator's argument, every term drops to a lower factorial and lands on a series you already know.

The target "known" series are the Maclaurin expansions at $x=1$: $e=\sum_{n=0}^{\infty}\frac{1}{n!},\qquad \cosh 1=\sum_{n=0}^{\infty}\frac{1}{(2n)!}=\frac{e+e^{-1}}{2},\qquad \sinh 1=\sum_{n=0}^{\infty}\frac{1}{(2n+1)!}=\frac{e-e^{-1}}{2}.$

Method

1. Identify the denominator's argument $m=an+b$ (e.g. $m=2n+1$). This is the quantity you will express the numerator in terms of.
2. Rewrite the numerator $P(n)$ as a linear combination of $1,\,m,\,m(m-1),\dots$ Solve for the constants by matching. For a linear $P$, write $n=\dfrac{1}{a}\big[(an+b)-b\big]$.
3. Cancel one factor of the argument per term: $\dfrac{m}{m!}=\dfrac{1}{(m-1)!}$, and $\dfrac{m(m-1)}{m!}=\dfrac{1}{(m-2)!}$, lowering each piece to a known factorial series.
4. Match each resulting sum to $e$, $\sinh 1$, or $\cosh 1$ (mind whether the index gives even or odd factorials), then combine.

Worked example

Evaluate $\displaystyle S=\sum_{n=0}^{\infty}\frac{n}{(2n+1)!}$.

Here the argument is $m=2n+1$, so write $n=\tfrac12\big[(2n+1)-1\big]$: $\frac{n}{(2n+1)!}=\frac12\cdot\frac{2n+1}{(2n+1)!}-\frac12\cdot\frac{1}{(2n+1)!}=\frac12\cdot\frac{1}{(2n)!}-\frac12\cdot\frac{1}{(2n+1)!}.$ Summing each piece against the known series, $S=\frac12\sum_{n=0}^{\infty}\frac{1}{(2n)!}-\frac12\sum_{n=0}^{\infty}\frac{1}{(2n+1)!}=\frac12\cosh 1-\frac12\sinh 1=\frac{1}{2}\cdot\frac{2e^{-1}}{2}=\frac{1}{2e}.$

The same recipe handles $\sum \dfrac{n}{n!}=e$, $\sum \dfrac{n^2}{n!}=2e$, and quadratic numerators (split $n^2=n(n-1)+n$ to cancel two factors).

Exam trick: Pick the split that matches the denominator's exact argument, not just any factor of $n$. For $(2n+1)!$ use $n=\tfrac12[(2n+1)-1]$ (giving $(2n)!$ and $(2n+1)!$); for $(2n)!$ use $n=\tfrac12[(2n)-0]+\dots$ so the cancelled factor leaves the parity you can sum. Cancelling the wrong factor lands you on a series you can't name.