Using the Cauchy criterion, test the uniform convergence
Using the Cauchy criterion, test the uniform convergence of the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{n^2}$ on $[-1,1]$.
Explanation of Method
Bounding $|x|\le1$ reduces the tail sum to $\sum_{k=m+1}^n1/k^2$, whose smallness is already guaranteed by the ordinary (numeric) Cauchy criterion for the convergent series $\sum1/k^2$ — no dependence on $x$ remains. This is one instance where the extra decay $1/n^2$ (versus $\sum x^n$ in Problem 184) is strong enough to secure uniform convergence all the way out to the boundary $|x|=1$.
- Bound the tail sum by $\sum_{k=m+1}^n1/k^2$, independent of $x$.
- Apply the numeric Cauchy criterion for the convergent series $\sum1/k^2$.
- Confirm via the Weierstrass M-test as a cross-check.
Step 1 · Bound the tail sum
For $x\in[-1,1]$ and $n>m$: $$ \left|\sum_{k=m+1}^n\frac{x^k}{k^2}\right| \le \sum_{k=m+1}^n\frac{|x|^k}{k^2} \le \sum_{k=m+1}^n\frac1{k^2}, $$ using $|x|\le1$.
Step 2 · Apply the numeric Cauchy criterion
Since $\sum1/k^2$ converges, given $\epsilon>0$ there is $N$ with $\sum_{k=m+1}^n1/k^2<\epsilon$ for all $n>m\ge N$. Since $\sum_{k=m+1}^{\infty}1/k^2<\int_m^{\infty}dt/t^2=1/m$, one can take $N>1/\epsilon$ explicitly. This bound is independent of $x$, so the Cauchy criterion holds uniformly on $[-1,1]$.
Step 3 · Weierstrass M-test cross-check
$|x^n/n^2|\le1/n^2=M_n$ for $x\in[-1,1]$, and $\sum M_n=\pi^2/6<\infty$, confirming uniform (and absolute) convergence.
Summary Table
| Bound | Value |
|---|---|
| Tail sum bound | $\le1/m$, independent of $x$ |
| Uniform convergence on $[-1,1]$ | Yes |
Additional Notes
- This is the direct positive counterpart to Problem 184's negative result: replacing the plain geometric series $\sum x^n$ with the $1/n^2$-weighted version restores uniform convergence all the way to the closed boundary $|x|=1$, exactly the same qualitative jump seen between Problems 163 (conditional at one endpoint) and 173/174 (fully closed boundary of absolute convergence).
- The endpoint values are meaningful here: at $x=1$, the series equals $\pi^2/6$; at $x=-1$, it equals $-\pi^2/12$ (an alternating $\eta$-function value) — both genuine, finite sums, consistent with the series being well-behaved right up to and including the boundary.
- This problem cleanly illustrates that the Cauchy criterion and the Weierstrass M-test, while conceptually different (one direct from the definition, one via a majorant), give identical conclusions whenever a clean $x$-independent majorant like $1/n^2$ is available.
Result
The series converges uniformly on $[-1,1]$: