Question on Uniform Conergence

Question

I need to show that $\sum_{k=1}^\infty$$(\frac {x}{2})^k$ does not converge uniformly on (-2, 2)

I know I have to show that $\sup_{x\in X}\lvert f_n(x)-f(x)\rvert\nrightarrow0 $ as $n\rightarrow\infty$, because if the sequence of partial sums does not converge uniformly then the series can't either.

but I need some help setting up $\sup_{x\in X}\lvert f_n(x)-f(x)\lvert $

Answer 1

Consider

$$\sum_{k=1}^n (x/2)^k.$$

Now pick $x=2-2/n$.

Answer 2

Suppose it did converge uniformly to a function $f$. Then there would exist an index $n$ with the property that $$\left| \sum_{k=1}^n \left( \frac x 2 \right)^k - f(x) \right| < 1$$ for all $x \in (-2,2)$. On the other hand, the finite sum is bounded on this interval but the limit function $f$ (which you can compute explicitly as a geometric series) is not bounded. Thus the difference cannot, for any $n$, be uniformly less than $1$.

Answer 3

Notice that the given sequence is a partial sum of a geometric sequence hence

$$\mu_n(x):=f(x)-f_n(x)=\sum_{k=n+1}^\infty(x/2)^k=(x/2)^{n+1}\frac{1}{1-x/2}$$ so we have that $$\sup_{x\in(-2,2)}|f(x)-f_n(x)|\ge\lim_{x\to2}\mu_n(x)=+\infty$$ and thus the sequence cannot converge uniformly.