To prove $\sum_{n=0}^{\infty}x^n$ does not unif. converge on $(-1,1)$, let $f_n$ be the partial sum, can I use the difference $|\frac{1}{1-x} - f_n|$ diverges?
Prove $\sum_{n=0}^{\infty}x^n$ not converge uniformly
0
$\begingroup$
real-analysis
sequences-and-series
uniform-convergence
-
0Yes, you can (and should). – 2017-02-13
1 Answers
1
We have $|\frac{1}{1-x}-\sum_{n=0}^N x^n|= \frac{|x|^{N+1}}{1-x} \to \infty$ for $x \to 1$. This shows that $(\sum_{n=0}^N x^n)_N$ does not converge uniformly to $\frac{1}{1-x}$ on $(-1,1)$.