Find the domain of definition of f(x)=sum from n=1
Find the domain of definition of $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{e^{-n(x+1)}}{n^2}$.
Explanation of Method
Setting $y=e^{-(x+1)}$ turns the series into $\sum y^n/n^2$, the dilogarithm series, which converges for $|y|\le1$ (the boundary included, since $\sum1/n^2$ and $\sum(-1)^n/n^2$ both converge absolutely). Since $y=e^{-(x+1)}$ is always positive, the condition $y\le1$ translates directly into a half-line condition on $x$.
- Substitute $y=e^{-(x+1)}$ and identify the dilogarithm series.
- Confirm convergence at $|y|=1$ (both endpoints).
- Translate $y\le1$ back into a condition on $x$.
Step 1 · Substitution
With $y=e^{-(x+1)}$: $$ f(x) = \sum_{n=1}^{\infty}\frac{y^n}{n^2} = \mathrm{Li}_2(y), $$ the dilogarithm function, whose defining series has radius of convergence $1$ in $y$ (ratio test: $\lim|a_{n+1}/a_n|=\lim(n/(n+1))^2=1$).
Step 2 · Boundary convergence
At $y=1$: $\sum1/n^2=\pi^2/6<\infty$, converges. At $y=-1$: $\sum(-1)^n/n^2$ converges absolutely (dominated by $\sum1/n^2$). So the series in $y$ converges for the full closed range $|y|\le1$.
Step 3 · Translate to $x$
Since $y=e^{-(x+1)}>0$ always, the condition $|y|\le1$ reduces to $y\le1$: $$ e^{-(x+1)}\le1 \iff -(x+1)\le0 \iff x\ge-1. $$ As $x\to\infty$, $y\to0$, always within range; there is no upper restriction.
Summary Table
| Property | Value |
|---|---|
| Closed form | $\mathrm{Li}_2(e^{-(x+1)})$ |
| $f(-1)$ | $\pi^2/6$ |
| Domain of definition | $[-1,\infty)$ |
Additional Notes
- Since the dilogarithm's series converges on the closed interval $|y|\le1$ (unlike a generic power series, which typically fails at least at one endpoint), this domain problem resolves cleanly into a closed half-line $[-1,\infty)$ rather than an open one.
- As $x\to\infty$, $f(x)\to0$ exponentially fast (each term shrinks like $e^{-n(x+1)}$), while at the left endpoint $x=-1$ the function attains its maximum value $\pi^2/6$ — the same Basel-sum constant recurring throughout this problem collection.
- This is structurally the same exponential-substitution technique as Problem 206's $\sum xe^{-nx}$, except here the coefficient decay ($1/n^2$ instead of a bare geometric term) is strong enough to secure convergence right up to and including the boundary.