If sum from n=1 to infinity|u_n(x)| converges uniformly
If $\displaystyle\sum_{n=1}^{\infty}|u_n(x)|$ converges uniformly on $[a,b]$, prove that $\displaystyle\sum_{n=1}^{\infty}u_n(x)$ converges uniformly on $[a,b]$.
Explanation of Method
This is the uniform-convergence analogue of "absolute convergence implies convergence." The proof is a direct application of the Cauchy criterion for uniform convergence combined with the triangle inequality — the same one-line argument that works pointwise transfers immediately once suprema are taken.
- State the Cauchy criterion for uniform convergence of $\sum|u_n(x)|$.
- Apply the triangle inequality to bound partial sums of $\sum u_n(x)$.
- Conclude the Cauchy criterion holds for $\sum u_n(x)$ as well.
Step 1 · Cauchy criterion for $\sum|u_n(x)|$
Since $\sum_{n=1}^{\infty}|u_n(x)|$ converges uniformly on $[a,b]$: for every $\epsilon>0$ there exists $N$ such that for all $n>m\ge N$ and all $x\in[a,b]$, $$ \sum_{k=m+1}^{n}|u_k(x)| < \epsilon. $$
Step 2 · Triangle inequality
For the same $N$, $n>m\ge N$, and every $x\in[a,b]$: $$ \left|\sum_{k=m+1}^{n}u_k(x)\right| \le \sum_{k=m+1}^{n}|u_k(x)| < \epsilon. $$
Step 3 · Conclude
This shows $\sum_{n=1}^{\infty}u_n(x)$ satisfies the Cauchy criterion for uniform convergence on $[a,b]$, so it converges uniformly there.
Summary Table
| Hypothesis | Conclusion |
|---|---|
| $\sum|u_n(x)|$ uniformly convergent on $[a,b]$ | $\sum u_n(x)$ uniformly convergent on $[a,b]$ |
Additional Notes
- This theorem is exactly what licenses every Weierstrass M-test conclusion in this batch (Problems 192–203, 208, 217–220): the M-test proves uniform convergence of $\sum|u_n(x)|$ via a majorant $M_n$, and this proposition then transfers that to uniform convergence of $\sum u_n(x)$ itself.
- The converse fails: uniform convergence does not imply absolute uniform convergence. Problem 201's series $\sum(-1)^n/(x+n)$ converges uniformly on $(0,\infty)$, but $\sum1/(x+n)$ does not even converge (it behaves like the divergent harmonic series for each fixed $x$).
- The proof pattern here — Cauchy criterion plus triangle inequality — is the uniform-convergence mirror of the elementary pointwise fact that $|\sum a_n|\le\sum|a_n|$; nothing about uniformity changes the logic, only that the bound $\epsilon$ must hold simultaneously for every $x$ in the domain.
Result
Proved via the Cauchy criterion and the triangle inequality: