UIC Master’s Exam- Fall 2007 R1

Problem Statement: Consider the power series


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


(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


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].


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!!!

This entry was posted in Absolute Convergence, Analysis, Conditional Convergence, M-Test, Math, Series, Series of Functions, Uniform Convergence. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s