## UIC Master’s Exam- Fall 2007 R1

Problem Statement: Consider the power series

$\sum\limits_{n=1}^{\infty}\dfrac{1}{2^nn}x^n$

(a) For which values of $x$ does it converge absolutely? Conditionally? (b) Show that on $[-1,1]$ the convergence is uniform.

Solutions:

(a) Claim: the series converges absolutely on $(-2,2)$ and conditionally on $[-2,2)$.

Consider $x\in(-2,2)$, then it follows that $\left|\dfrac{x}{2}\right|<1$ and so we may find a rational $r\in(\dfrac{x}{2},1)$. Then it follows that $\left|\left(\dfrac{x}{2}\right)^n\dfrac{1}{n}\right|\leq\left|\left(\dfrac{x}{2}\right)^n\right|$ for every $n\in\mathbb{N}$. By the geometric series test $\sum\limits_{n=1}^{\infty}\left|\left(\dfrac{x}{2}\right)^n\right|$ converges and so by the comparison test $\sum\limits_{n=1}^{\infty}\left|\left(\dfrac{x}{2}\right)^n\dfrac{1}{n}\right|$ converges. Thus, the series converges absolutely on $(-2,2)$.

When $x=-2$ then the series becomes $\sum\limits_{n=1}^{\infty}(-1)^n\dfrac{1}{n}$, the alternating harmonic series, which we know converges by the AST. But when $x=2$ the series is simply the harmonic series which we know diverges. Thus, the series converges conditionally on $[-2,2)$.

(b) Proof: Consider the series above for $x\in[-1,1]$ it follows that

$\left|\dfrac{1}{2^nn}x^n\right|\leq\left|\dfrac{1}{2^nn}\right|\leq\left|\dfrac{1}{2^n}\right|$

for all $n\in\mathbb{N}$ and for all $x\in[-1,1]$

We know that $\sum\limits_{n=1}^{\infty}\dfrac{1}{2^n}$ converges by the p-series test (since $\dfrac{1}{2}<1$).  And so applying the M-Test we may conclude that $\sum\limits_{n=1}^{\infty}\dfrac{1}{2^nn}x^n$ converges uniformly on $[-1,1]$.

$\Box$

Reflection: This was a great reminder problem. Whenever I see uniform convergence my first thought is usually to try using the definition, which can be a real pain in the butt. This problem reminded me that when you’re dealing with uniform convergence you should try to use the M-Test if you can because it is usually much much simpler than trying to go straight from the definition. Try the easy way first!!!