By term-by-term integration, find the sum sum from n=0
By term-by-term integration, find the sum $\displaystyle\sum_{n=0}^{\infty}\frac{(2n+3)x^{2n}}{n!}$.
Explanation of Method
Splitting $(2n+3)$ into $2n$ and $3$ separates the sum into a shifted copy of $e^{x^2}$ (from the $2n$ part) and a plain multiple of $e^{x^2}$; the shifted piece is most easily obtained by differentiating $e^{x^2}$.
- Split into $2\sum nx^{2n}/n! + 3\sum x^{2n}/n!$.
- Evaluate the second sum directly, and the first via reindexing or differentiation.
- Combine and verify numerically at $x=1$.
Step 1 · Split the sum
$$ \sum_{n=0}^{\infty}\frac{(2n+3)x^{2n}}{n!} = 2\sum_{n=0}^{\infty}\frac{nx^{2n}}{n!} + 3\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. $$ The second sum is $3e^{x^2}$.
Step 2 · Evaluate the first sum
For $n\ge1$, $\dfrac{nx^{2n}}{n!}=\dfrac{x^{2n}}{(n-1)!}$; with $m=n-1$: $$ \sum_{n=1}^{\infty}\frac{x^{2n}}{(n-1)!} = \sum_{m=0}^{\infty}\frac{x^{2m+2}}{m!} = x^2\sum_{m=0}^{\infty}\frac{x^{2m}}{m!} = x^2e^{x^2}. $$
Step 3 · Combine and verify
$$ \sum_{n=0}^{\infty}\frac{(2n+3)x^{2n}}{n!} = 2x^2e^{x^2}+3e^{x^2} = (2x^2+3)e^{x^2}. $$ At $x=0$: gives $3$, matching the constant term. At $x=1$: gives $5e\approx13.591$, matching partial sums ($S_4\approx13.458$, converging toward $13.591$).
Summary Table
| Property | Value |
|---|---|
| Sum | $(2x^2+3)e^{x^2}$ |
| Region of validity | $\mathbb R$ |
| At $x=1$ | $5e\approx13.591$ |
Additional Notes
- Equivalently, $(2x^2+3)e^{x^2}=2x\cdot\frac{d}{dx}[e^{x^2}]+3e^{x^2}$, since $\frac{d}{dx}e^{x^2}=2xe^{x^2}$ — differentiating the base series and multiplying by $x$ is an alternative route to the same reindexing result.
- This is structurally the $x^2$-substituted cousin of Problem 292's $\sum(n+1)2^n/n!$: both split a linear-in-$n$ weight into a shifted exponential piece plus a plain exponential piece.
- Since $e^{x^2}$ is entire, this series converges for every real $x$, with no radius-of-convergence restriction to track.