Uniform Convergence of The series $\sum\limits_{i=n}^\mathbb{N}{x^n}$ on (-1,1)

Question

Question: Study the Convergence of $S_n=\sum\limits_{i=n}^\mathbb{N}{f_n}$ such that $f_n: I=(-1,1) \to \mathbb{R} \quad,\quad f_n(x)=x^n.$

We have proved that the series converges Pointwise to $S:(-1,1) \to \mathbb{R} \quad,\quad S(x)=\frac{1}{1-x}.$

Now we have to study the uniform Convergence, How to check if this series is uniform Convergent or Not ??

Thanks for Help.

Answer 1

1. Let $[a,b]$ a compact intervall contained in $I$. Then we find $q \in (0,1)$ such that

$-1<-q \le a < b \le q <1$.

If $x \in [a,b]$, then $|x| \le q$, hence $|x^n|=|x|^n \le q^n$. The Weierstrass test shows now that the geometric series $\sum_{n=0}^{\infty}x^n$ converges uniformly on $[a,b]$.

2. For $x \in I$ we have, with $S_n(x)=\sum_{k=0}^{n}x^k$ ,

   $$B_n(x):=|S_n(x)-S(x)|=\frac{|x|^{n+1}}{1-x}.$$

If $n $ is fixed, Then $B_n(x) \to \infty$ for $x \to 1$.

Hence, the series $\sum_{n=0}^{\infty}x^n$ does not converge uniformly on $I$.

Answer 2

The sequence of functions $f_n=x^n$ does not converges uniformly in $(-1,1)$ since $\lim |x^n|_{(-1,1)}=1 \neq 0$ (here $|\thinspace|_{(-1,1)}$ denotes the sup norm), hence $\sum f_n$ does not converge uniformly in $(-1,1)$.

However, if we fix $0<\delta < 1$, we can prove $\sum f_n$ converges uniformly in $[-\delta, \delta]$. In this interval, we have $$ |x^n|=|x|^n \leq \delta^n.$$ Let $\epsilon > 0$. Since $\sum \delta^n$ converges (as $\delta < 1$), there exist $N \in \mathbb{N}$ such that for all $n>m \geq N$ we have $$ \Big|\sum_{m+1}^{n} x^n\Big|\leq \sum_{m+1}^{n} |x|^n \leq \sum_{m+1}^{n} \delta^n<\epsilon.$$ So, $\sum f_n$ is Cauchy and hence converges uniformly in $[-\delta, \delta]$.

Answer 3

The series is not uniformly convergent on $I = (-1,1)$. Let $A_n(x)$be the $n$th partial sum of $\sum_{m=0}^{\infty} x^m = \frac{1}{1-x}$. If the sum is uniformly convergent on $I$, then for any $\epsilon > 0$ there exists $N$ so that $|\frac{1}{1-x} -A_N(x)| < \epsilon$ for all $x \in I$. But $|\frac{1}{1-x}-A_n(x)| = |x^{N+1} + x^{N+2}+...| = |\frac{x^{N+1}}{1-x}|$. This goes to infinity as $x\to1^-$, so it exceeds $\epsilon$ for some $x$, which contradicts our assumption. It follows that the sum is not uniformly convergent. However, it does converge uniformly in compact subsets of $I$, as the other answer shows.