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!!!

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

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