Test the uniform convergence of sum from n=1 to infinity
Test the uniform convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{x+n}$ on $x\in(0,\infty)$.
Explanation of Method
The Weierstrass M-test fails here (the natural bound $1/(x+n)\to1/n$ as $x\to0^+$, and $\sum1/n$ diverges). But the series is alternating with terms $a_n(x)=1/(x+n)$ decreasing in $n$ to $0$ for every fixed $x>0$, so the Leibniz remainder bound applies — and crucially, that bound is itself uniform in $x$, which is enough to prove uniform convergence directly, without needing the M-test at all.
- Confirm pointwise convergence via the Alternating Series Test.
- Bound the remainder using the Leibniz estimate, uniformly in $x$.
- Conclude uniform convergence directly from that $x$-independent bound.
Step 1 · Pointwise convergence
Fix $x>0$ and let $a_n=1/(x+n)$. Then $a_n>0$, $a_n$ is strictly decreasing in $n$, and $a_n\to0$. By the Leibniz (Alternating Series) Test, $\sum(-1)^na_n$ converges for every $x>0$.
Step 2 · A uniform remainder bound
For an alternating series with terms decreasing to $0$, the remainder after $N$ terms satisfies the standard estimate $$ |R_N(x)| = \left|\sum_{n=N+1}^{\infty}\frac{(-1)^n}{x+n}\right| \le a_{N+1}(x) = \frac1{x+N+1}. $$ Since $x>0$: $x+N+1>N+1$, so $$ |R_N(x)| < \frac1{N+1} $$ for every $x\in(0,\infty)$ — the bound $1/(N+1)$ does not depend on $x$ at all.
Step 3 · Uniform convergence
Taking the supremum over $x>0$: $$ \sup_{x>0}|R_N(x)| \le \frac1{N+1} \longrightarrow 0 \text{ as } N\to\infty. $$ This is precisely the definition of uniform convergence: given $\epsilon>0$, choose $N$ with $1/(N+1)<\epsilon$; then $|R_n(x)|<\epsilon$ for every $n\ge N$ and every $x\in(0,\infty)$ simultaneously. So the series converges uniformly on $(0,\infty)$ (and the same bound extends this to $[0,\infty)$).
Summary Table
| Test | Outcome |
|---|---|
| Weierstrass M-test | Fails ($\sum1/n$ diverges) |
| Uniform Leibniz remainder bound | $|R_N(x)|\le1/(N+1)$, works |
| Uniform convergence on $(0,\infty)$ | Yes |
Additional Notes
- This is the classic textbook illustration that the Weierstrass M-test is only a sufficient condition: it fails here, yet uniform convergence holds anyway — the alternating structure supplies its own $x$-independent error bound, exactly the mechanism used later in Problems 205 and 217 (Abel's and Dirichlet's tests) to rescue series where the M-test cannot apply.
- The key algebraic fact driving the proof is trivial but essential: $x+N+1>N+1$ whenever $x>0$, so a bound that would only hold "in the limit as $x\to0^+$" for the raw term $1/(x+n)$ becomes a genuine, always-true inequality for the remainder bound $1/(x+N+1)<1/(N+1)$.
- Contrast this with Problem 195's series ∑nx/(1+n⁴x²), where the analogous Cauchy-criterion check genuinely fails near the troublesome point — here, the alternating sign is what saves the series from that fate.
Result
Converges uniformly on $(0,\infty)$ (Weierstrass M-test fails, but the uniform Leibniz bound succeeds):