If sum from n=1 to infinity a_n converges, prove that sum
If $\displaystyle\sum_{n=1}^{\infty}a_n$ converges, prove that $\displaystyle\sum_{n=1}^{\infty}a_ne^{-nx}$ converges uniformly on $x\in[0,\infty)$.
Explanation of Method
This is precisely the setting of Abel's test for uniform convergence: a fixed convergent series $\sum a_n$ (which is trivially "uniform" in $x$, since it doesn't depend on $x$ at all) multiplied by a factor $b_n(x)=e^{-nx}$ that is uniformly bounded ($0<b_n(x)\le1$) and monotone decreasing in $n$ for every fixed $x\ge0$. Abel's summation-by-parts lemma converts the Cauchy tail bound on $\sum a_n$ into a uniform Cauchy tail bound on $\sum a_ne^{-nx}$ — valid on the entire half-line $[0,\infty)$, including $x=0$.
- State Abel's summation lemma for a monotone bounded factor.
- Verify $b_n(x)=e^{-nx}$ is uniformly bounded and monotone in $n$ for each $x\ge0$.
- Combine with the Cauchy criterion for $\sum a_n$ to get uniform convergence on all of $[0,\infty)$.
Step 1 · Abel's summation lemma
If $b_{m+1}\ge b_{m+2}\ge\cdots\ge b_n\ge0$ and $\left|\sum_{k=m+1}^{j}a_k\right|\le\epsilon$ for every $j\in\{m+1,\dots,n\}$, then by summation by parts: $$ \left|\sum_{k=m+1}^{n}a_kb_k\right| \le \epsilon\, b_{m+1}. $$
Step 2 · Verify the hypotheses on $b_n(x)=e^{-nx}$
For each fixed $x\ge0$: $b_{n+1}(x)/b_n(x)=e^{-x}\le1$, so $b_n(x)$ is decreasing in $n$; and $b_n(x)=e^{-nx}\in(0,1]$ for all $n\ge1,\,x\ge0$, so $b_{m+1}(x)\le1$ — a bound independent of $x$.
Step 3 · Apply Abel's lemma uniformly
Since $\sum a_n$ converges, by the Cauchy criterion, for every $\epsilon>0$ there is $N$ such that $\left|\sum_{k=m+1}^{j}a_k\right|<\epsilon$ for all $n>m\ge N$ and all $j\in\{m+1,\dots,n\}$. By Step 1's lemma (applied with $b_k=e^{-kx}$ for each fixed $x$): $$ \left|\sum_{k=m+1}^{n}a_ke^{-kx}\right| \le \epsilon\cdot e^{-(m+1)x} \le \epsilon $$ for every $x\ge0$ simultaneously (using $e^{-(m+1)x}\le1$). This is the Cauchy criterion for uniform convergence, so $\sum a_ne^{-nx}$ converges uniformly on $[0,\infty)$.
Summary Table
| Component | Role |
|---|---|
| $\sum a_n$ convergent | Supplies the uniform-in-$x$ tail bound $\epsilon$ |
| $b_n(x)=e^{-nx}$ | Uniformly bounded ($\le1$), monotone decreasing in $n$ |
| Conclusion | $\sum a_ne^{-nx}$ uniformly convergent on all of $[0,\infty)$ |
Additional Notes
- Unlike a naive attempt to bound $|a_ne^{-nx}|\le|a_n|$ (which would need $\sum a_n$ to converge absolutely for the Weierstrass M-test), Abel's test needs only ordinary convergence of $\sum a_n$ — the monotone bounded factor $e^{-nx}$ does all the extra work via summation by parts, exactly as in the sharper Hilbert-inequality proofs of Problems 160/162.
- Uniform convergence holds on the entire closed half-line $[0,\infty)$, including $x=0$ where the series reduces to $\sum a_n$ itself — there is no need to restrict to $x\ge\delta>0$, since the bound $e^{-(m+1)x}\le1$ already holds uniformly at every point of the domain, including the boundary.
- This is the exponential-weight analogue of Problem 201's alternating-series result: both problems show a series that fails the Weierstrass M-test can still converge uniformly, once a monotone auxiliary factor (alternating sign there, exponential decay here) is exploited via Abel/Dirichlet-type summation-by-parts arguments.
Result
By Abel's test, uniform convergence holds on the entire domain $[0,\infty)$: