Using the Cauchy criterion, test the uniform convergence
Using the Cauchy criterion, test the uniform convergence of the series $\displaystyle\sum_{n=0}^{\infty}x^n$ for: (1) $|x|<q<1$; (2) $|x|<1$.
Explanation of Method
The Cauchy criterion requires the tail sums $\left|\sum_{k=m+1}^nx^k\right|$ to be small uniformly in $x$, not just for each fixed $x$. On $|x|\le q<1$, the geometric tail is bounded by $2q^{m+1}/(1-q)$, independent of $x$ and $\to0$. On the full open disk $|x|<1$, the factor $1/(1-x)$ blows up as $x\to1^-$, so no single bound works for all $x$ simultaneously — the criterion fails.
- Bound the tail sum on $|x|\le q<1$, independent of $x$.
- Show the Cauchy criterion fails on $|x|<1$ by letting $x\to1^-$.
Step 1 · Uniform bound for $|x|\le q<1$
$$ \left|\sum_{k=m+1}^nx^k\right| = \left|\frac{x^{m+1}-x^{n+1}}{1-x}\right| \le \frac{|x|^{m+1}+|x|^{n+1}}{1-|x|} \le \frac{2q^{m+1}}{1-q}, $$ using $|1-x|\ge1-|x|\ge1-q$ and $|x|\le q$. Given $\epsilon>0$, choosing $N$ with $2q^{N+1}/(1-q)<\epsilon$ makes the Cauchy criterion hold uniformly for all $n>m\ge N$ and all $|x|\le q$. Uniform convergence holds.
Step 2 · Failure on $|x|<1$
For any fixed $m$, choose $x=1-\dfrac1{m+1}$ (so $|x|<1$). Then $|x|^{m+1}=\left(1-\frac1{m+1}\right)^{m+1}\to e^{-1}$, while $1-x=\dfrac1{m+1}\to0$, so $$ \frac{|x|^{m+1}}{1-x} \approx (m+1)e^{-1} \to \infty. $$ So for any $N$, one can find $m\ge N$ and $x$ with $|x|<1$ making the tail sum arbitrarily large — the Cauchy criterion fails uniformly. No uniform convergence on $|x|<1$.
Summary Table
| Domain | Uniform convergence? |
|---|---|
| $|x|\le q<1$ (any fixed $q$) | Yes |
| $|x|<1$ (full open disk) | No |
Additional Notes
- This is the geometric-series prototype for the general phenomenon seen throughout this batch: uniform convergence typically holds on any closed sub-region strictly inside the natural domain of convergence, but fails on the full open domain whenever the sum itself blows up approaching the boundary.
- The bound $2q^{m+1}/(1-q)$ is the direct analogue of the Weierstrass M-test bound $M_n=q^n$ (with $\sum M_n=q^{m+1}/(1-q)<\infty$), so this problem is really a Cauchy-criterion restatement of what the Weierstrass test would give immediately for part (1).
- Part (2)'s failure is essentially the same mechanism as Problem 199's failure of uniform convergence for $\sum1/n^x$ on $(1,\infty)$: both series remain pointwise convergent throughout the open domain but blow up in a way that defeats any single uniform bound as the boundary is approached.
Result
Uniformly convergent on $|x|\le q<1$; not uniformly convergent on $|x|<1$: