Using the Weierstrass test, show that the series sum
Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac1{n^x}$ converges uniformly on $[a,\infty)$ for any fixed $a>1$, but not uniformly on $(1,\infty)$ itself.
Explanation of Method
For $x\ge a>1$, $n^x\ge n^a$, so each term is bounded by $1/n^a$, and $\sum1/n^a=\zeta(a)$ converges — the M-test applies on $[a,\infty)$. But as $a\to1^+$, $\zeta(a)\to\infty$, so no single convergent majorant works on all of $(1,\infty)$; a direct Cauchy-criterion argument confirms uniform convergence genuinely fails there.
- Bound each term by $1/n^a$ on $[a,\infty)$ and apply the M-test.
- Show uniform convergence fails on $(1,\infty)$ via the Cauchy criterion near $x=1^+$.
- Identify the sum on $[a,\infty)$ as the Riemann zeta function.
Step 1 · Uniform convergence on $[a,\infty)$, $a>1$
For $x\ge a$ and $n\ge1$: $n^x\ge n^a$ (since $n\ge1$), so $\dfrac1{n^x}\le\dfrac1{n^a}$. Since $\sum1/n^a=\zeta(a)<\infty$ for $a>1$, the Weierstrass M-test with $M_n=1/n^a$ gives uniform convergence on $[a,\infty)$.
Step 2 · Failure on $(1,\infty)$ itself
Fix any $m<k$. As $x\to1^+$: $$ \sum_{n=m}^{k}\frac1{n^x} \longrightarrow \sum_{n=m}^{k}\frac1n, $$ which can be made arbitrarily large by choosing $k$ large (harmonic tail sums are unbounded). So for any $x$ close enough to $1$, the partial tail sum $\sum_{n=m}^{k}1/n^x$ exceeds any fixed threshold — the Cauchy criterion for uniform convergence fails as $x$ ranges over all of $(1,\infty)$, since points near $x=1$ always spoil it regardless of $m$.
Step 3 · The zeta function
On $[a,\infty)$, the uniform limit $\displaystyle\zeta(x)=\sum_{n=1}^{\infty}\frac1{n^x}$ is therefore continuous on $[a,\infty)$ for every $a>1$, hence continuous on all of $(1,\infty)$ (continuity is a local property, so uniform convergence on each closed sub-interval $[a,\infty)$ suffices to conclude continuity on the open interval).
Summary Table
| Domain | Uniform convergence? |
|---|---|
| $[a,\infty)$, any fixed $a>1$ | Yes ($M_n=1/n^a$, $\sum M_n=\zeta(a)$) |
| $(1,\infty)$ | No (fails near $x=1^+$) |
Additional Notes
- This is the Dirichlet-series analogue of Problem 195's phenomenon: convergence is uniform on any region bounded away from the "bad" boundary point ($x=1$ here, $x=0$ there), but fails on the full open domain that includes it — a recurring pattern across this batch (also seen in Problems 176, 181, 182, 187).
- The failure at $x=1$ is exactly the divergence of the harmonic series $\sum1/n$, so this problem is really a uniform-convergence refinement of the elementary fact that $\zeta(x)\to\infty$ as $x\to1^+$.
- Via Problem 191's equivalence theorem, this same uniform-convergence-on-$[a,\infty)$-but-not-$(1,\infty)$ result transfers directly to the corresponding Newton's series $\sum a_n\binom{x-1}{n}$-type expression with $a_n=1$, giving a concrete instance of that general correspondence.
Result
Uniformly convergent on every $[a,\infty)$ with $a>1$, but not on $(1,\infty)$ itself: