Determine the domain of definition and the domain
Determine the domain of definition and the domain of continuity of $\displaystyle f(x)=\sum_{n=1}^{\infty}xe^{-nx}$.
Explanation of Method
For fixed $x\ne0$, the series is $x$ times a geometric series in $e^{-x}$, giving a closed form $x/(e^x-1)$ wherever it converges. Checking convergence directly at $x=0$ and $x<0$ pins down the domain; comparing the closed form's limit at $x\to0^+$ against the directly-computed value $f(0)=0$ reveals a removable discontinuity there.
- Sum the geometric series for $x>0$ to get a closed form.
- Check convergence at $x=0$ and divergence for $x<0$ to find the domain.
- Compare $\lim_{x\to0^+}f(x)$ with $f(0)$ to determine continuity.
Step 1 · Closed form for $x>0$
For $x>0$, $e^{-x}\in(0,1)$, so $\sum_{n=1}^{\infty}e^{-nx}=\dfrac{e^{-x}}{1-e^{-x}}=\dfrac1{e^x-1}$ is a convergent geometric series. Hence $$ f(x) = x\sum_{n=1}^{\infty}e^{-nx} = \frac{x}{e^x-1}, \qquad x>0. $$
Step 2 · The domain
At $x=0$: every term is $0\cdot e^0=0$, so $f(0)=0$ and the series converges. For $x<0$: $e^{-nx}=e^{n|x|}\to\infty$, so the terms do not tend to $0$ and the series diverges. Hence the domain of definition is $[0,\infty)$.
Step 3 · Continuity at $x=0$
Using $e^x-1=x\left(1+\dfrac x2+\dfrac{x^2}6+\cdots\right)$: $$ \lim_{x\to0^+}\frac{x}{e^x-1} = \lim_{x\to0^+}\frac1{1+\frac x2+\cdots} = 1. $$ Since $f(0)=0\ne1=\lim_{x\to0^+}f(x)$, $f$ has a jump discontinuity at $x=0$ (removable if $f(0)$ were redefined to be $1$). For $x>0$, $f(x)=x/(e^x-1)$ is a quotient of continuous functions with nonvanishing denominator, hence continuous there.
Summary Table
| Property | Value |
|---|---|
| Domain of definition | $[0,\infty)$ |
| Closed form ($x>0$) | $f(x)=x/(e^x-1)$ |
| $f(0)$ | $0$ |
| $\lim_{x\to0^+}f(x)$ | $1$ |
| Domain of continuity | $(0,\infty)$ |
Additional Notes
- This discontinuity at $x=0$ is not surprising in light of Problem 202: the series $\sum u_n(x)$ with $u_n(x)=xe^{-nx}$ fails to converge uniformly near $x=0$ (each term individually vanishes there while the closed-form limit does not), a signature of exactly the kind of boundary breakdown catalogued across this batch.
- Redefining $f(0)=1$ removes the discontinuity entirely, since $x/(e^x-1)\to1$ smoothly as $x\to0$ — this is the same removable-singularity phenomenon familiar from $\sin x/x\to1$, and reflects that the series representation, not the underlying analytic function, is responsible for the jump.
- Compare with Problem 208, where the analogous series $\sum1/(n^2(1+n^2x^2))$ enjoys full uniform convergence and continuity on all of $\mathbb R$ — the presence of the factor $x$ multiplying a geometric (rather than power-law) tail is what causes Problem 206's series to break down precisely at the one point $x=0$.
Result
Defined on $[0,\infty)$, continuous only on $(0,\infty)$: