Using the Cauchy criterion, test the uniform convergence

Using the Cauchy criterion, test the uniform convergence of the series $\displaystyle\sum_{n=0}^{\infty}x(1-x)^n$ on $[0,1]$.

Step 1 · Closed-form partial sum and pointwise limit

For $x>0$: $S_N(x)=x\displaystyle\sum_{n=0}^N(1-x)^n=x\cdot\dfrac{1-(1-x)^{N+1}}x=1-(1-x)^{N+1}\to1$. At $x=0$: every term is $0$, so $S_N(0)=0$. So $$ S(x)=\begin{cases}0,&x=0\\1,&0<x\le1.\end{cases} $$

Step 2 · Remainder

For $x>0$: $|S(x)-S_N(x)|=(1-x)^{N+1}$. As $x\to0^+$, this $\to1$, so $\sup_{x\in[0,1]}|S(x)-S_N(x)|=1$ for every $N$ (approached, not attained).

Step 3 · Conclusion

$\lim_N\sup_{[0,1]}|S-S_N|=1\ne0$: not uniform. Equivalently: each partial sum $S_N$ is continuous, but $S$ jumps at $x=0$ — a uniform limit of continuous functions must be continuous, so the convergence cannot be uniform.

Summary Table

Additional Notes

• This series is the "series version" of Problem 176's sequence $x^n$: the partial sums $S_N(x)=1-(1-x)^{N+1}$ are themselves a disguised copy of $1-y^{N+1}$ with $y=1-x$, so the same discontinuous-limit mechanism recurs.
• The Cauchy-criterion computation directly confirms this: for $n>m$, $\left|\sum_{k=m+1}^nx(1-x)^k\right|=(1-x)^{m+1}\left[1-(1-x)^{n-m}\right]$, which approaches $1$ as $x\to0^+$ regardless of how large $m$ is — the criterion fails at every stage.
• This is a useful reminder that summing a family of otherwise well-behaved (even positive, even smooth) functions can still produce a discontinuous limit, precisely because the convergence is not uniform near the problematic point.