Find the sum sum from n=0 to infinity (2^n(n+1))/(n!)
Find the sum $\displaystyle\sum_{n=0}^{\infty}\frac{2^n(n+1)}{n!}$.
Explanation of Method
Splitting $(n+1)$ into $n$ plus $1$ separates the sum into a shifted exponential series (from the $n$ part) and a plain exponential series; both evaluate at $x=2$ to multiples of $e^2$.
- Split into $\sum n\cdot2^n/n! + \sum2^n/n!$.
- Reindex the first sum to reveal another copy of $e^2$.
- Combine and cross-check via differentiating $e^x$.
Step 1 · Split the sum
$$ \sum_{n=0}^{\infty}\frac{2^n(n+1)}{n!} = \sum_{n=0}^{\infty}\frac{n\cdot2^n}{n!} + \sum_{n=0}^{\infty}\frac{2^n}{n!}. $$ The second sum is immediately $e^2$.
Step 2 · Reindex the first sum
Since $\dfrac{n\cdot2^n}{n!}=\dfrac{2^n}{(n-1)!}$ for $n\ge1$, letting $m=n-1$: $$ \sum_{n=1}^{\infty}\frac{2^n}{(n-1)!} = \sum_{m=0}^{\infty}\frac{2^{m+1}}{m!} = 2\sum_{m=0}^{\infty}\frac{2^m}{m!} = 2e^2. $$
Step 3 · Combine and cross-check
$$ \sum_{n=0}^{\infty}\frac{2^n(n+1)}{n!} = 2e^2+e^2 = 3e^2 \approx 22.1672. $$ Cross-check via differentiation: from $\sum nx^n/n!=xe^x$, adding $e^x$ gives $\sum(n+1)x^n/n!=(x+1)e^x$; at $x=2$ this is $3e^2$, matching. Partial sums confirm: $S_6\approx21.889$, $S_{10}\approx22.165$, approaching $3e^2\approx22.167$.
Summary Table
| Property | Value |
|---|---|
| General formula | $\sum(n+k)x^n/n!=(x+k)e^x$ |
| Sum | $3e^2\approx22.1672$ |
Additional Notes
- The general pattern $\sum_{n=0}^\infty(n+k)x^n/n!=(x+k)e^x$ subsumes both this problem ($x=2,k=1$) and Problem 291 ($x=1$, after the $n(n-1)+n$ decomposition) as special cases.
- Splitting $(n+1)=n+1$ before summing is simpler here than the $n(n-1)+n$ trick needed for Problem 291's $n^2$, since $(n+1)$ is already linear in $n$.
- As with all these exponential-weighted sums, convergence is extremely fast (factorial in the denominator), so only about 10 terms are needed to match $3e^2$ to three decimal places.