Find the sum sum from n=1 to infinity frac nn+2·(x^n)/(n!)
Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac n{n+2}\cdot\frac{x^n}{n!}$.
Explanation of Method
Rewriting $\frac n{n+2}=1-\frac2{n+2}$ splits the sum into $e^x-1$ minus a piece whose denominator $(n+2)\cdot n!$ suggests reindexing through $(n+2)!$; carefully separating that piece into two shifted exponential series and simplifying yields a closed form in $e^x$ and $x$.
- Split using $n/(n+2)=1-2/(n+2)$.
- Reindex the resulting $1/[(n+2)\cdot n!]$ series through shifted exponentials.
- Combine and verify against the low-order series terms.
Step 1 · Split the sum
$$ \sum_{n=1}^{\infty}\frac n{n+2}\cdot\frac{x^n}{n!} = \sum_{n=1}^{\infty}\frac{x^n}{n!} - 2\sum_{n=1}^{\infty}\frac{x^n}{(n+2)\cdot n!} = (e^x-1) - 2\sum_{n=1}^{\infty}\frac{x^n}{(n+2)\cdot n!}. $$
Step 2 · Reindex the second sum
Since $(n+2)\cdot n!$ relates to $(n+2)!=(n+2)(n+1)n!$, write $\dfrac1{(n+2)n!}=\dfrac{n+1}{(n+2)!}$, then with $k=n+2$: $$ \sum_{n=1}^{\infty}\frac{(n+1)x^n}{(n+2)!} = \frac1{x^2}\sum_{k=3}^{\infty}\frac{(k-1)x^k}{k!} = \frac1{x^2}\left[(x-1)e^x+1-\frac{x^2}2\right], $$ using $\sum_{k=0}^\infty kx^k/k!=xe^x$ and $\sum_{k=0}^\infty x^k/k!=e^x$ with the $k=0,1,2$ terms subtracted off.
Step 3 · Combine and verify
$$ \sum_{n=1}^{\infty}\frac n{n+2}\cdot\frac{x^n}{n!} = (e^x-1) - \frac2{x^2}\left[(x-1)e^x+1-\frac{x^2}2\right] = \frac{(x^2-2x+2)e^x-2}{x^2}. $$ Expanding the right side in powers of $x$ reproduces the original series exactly: $\frac x3+\frac{x^2}4+\frac{x^3}{10}+\frac{x^4}{36}+\cdots$. At $x=1$: gives $e-2\approx0.718$.
Summary Table
| Property | Value |
|---|---|
| Sum | $[(x^2-2x+2)e^x-2]/x^2$ |
| Region of validity | $x\ne0$ |
| At $x=1$ | $e-2\approx0.718$ |
Additional Notes
- Although the formula has an apparent singularity at $x=0$, the original series is perfectly well-defined there (every term vanishes), so the closed form's $x\to0$ limit must — and does — equal $0$, since the numerator $(x^2-2x+2)e^x-2$ vanishes to second order at $x=0$.
- This kind of "coefficient is a rational function of $n$" sum always reduces to a combination of $e^x$ and low-degree polynomials in $x$, obtained by writing the rational function as a sum of shifted-factorial pieces, exactly as $n/(n+2)$ was split here.
- Verifying a closed form by re-expanding it in powers of $x$ and matching against the first several terms of the original series (as done in Step 3) is the standard way to catch algebraic slips in this style of derivation.