Find the sum sum from n=1 to infinity (n²)/(n!)
Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{n!}$.
Explanation of Method
Writing $n^2=n(n-1)+n$ splits each term into two pieces that telescope neatly against the exponential series after shifting the summation index — no differentiation required.
- Decompose $n^2=n(n-1)+n$.
- Reindex each resulting sum to recover the plain exponential series.
- Add the two results and verify numerically.
Step 1 · Decompose the numerator
$$ \sum_{n=1}^{\infty}\frac{n^2}{n!} = \sum_{n=1}^{\infty}\frac{n(n-1)}{n!} + \sum_{n=1}^{\infty}\frac{n}{n!} = \sum_{n=2}^{\infty}\frac{n(n-1)}{n!} + \sum_{n=1}^{\infty}\frac{n}{n!}. $$
Step 2 · Reindex each piece
Since $\dfrac{n(n-1)}{n!}=\dfrac1{(n-2)!}$ and $\dfrac n{n!}=\dfrac1{(n-1)!}$: $$ \sum_{n=2}^{\infty}\frac1{(n-2)!} = \sum_{k=0}^{\infty}\frac1{k!}=e, \qquad \sum_{n=1}^{\infty}\frac1{(n-1)!} = \sum_{m=0}^{\infty}\frac1{m!}=e. $$
Step 3 · Combine and verify
$$ \sum_{n=1}^{\infty}\frac{n^2}{n!} = e+e = 2e \approx 5.43656. $$ Partial sums confirm rapid convergence: $S_1=1$, $S_3=4.5$, $S_6=5.425$, $S_{10}\approx5.436544$, matching $2e$.
Summary Table
| Property | Value |
|---|---|
| Decomposition | $n^2=n(n-1)+n$ |
| Sum | $2e\approx5.4366$ |
Additional Notes
- The same result follows from differentiating $e^x=\sum x^n/n!$ twice (multiplying by $x$ between differentiations) and setting $x=1$ — a useful cross-check technique whenever a polynomial-in-$n$ numerator appears over $n!$.
- The decomposition $n^2=n(n-1)+n$ generalizes: any polynomial in $n$ can be rewritten in terms of falling factorials $n(n-1)(n-2)\cdots$, each of which cancels cleanly against $n!$ and reindexes to a fresh copy of $e$.
- This is the combinatorial partner to Problem 292's $\sum2^n(n+1)/n!=3e^2$ — both rely on the same falling-factorial trick, just evaluated at $x=1$ versus $x=2$.