## Saturday, May 5, 2007

Problem: Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 \cdot 2^n}$.

Solution: Let $\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$. Then $\displaystyle f^{\prime}(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = -\frac{\ln{(1-x)}}{x}$, a well-known Taylor series. So we want to integrate this:

$\displaystyle \int \frac{\ln{(1-x)}}{x}dx = \ln{x} \ln{(1-x)}-\int \frac{\ln{x}}{x-1}dx$

by parts using $u = \ln{(1-x)}$ and $dv = dx/x$. Substituting $t = 1-x$ in the last integral, we have

$\displaystyle \int \frac{\ln{(1-x)}}{x}dx = \ln{x} \ln{(1-x)}-\int \frac{\ln{(1-t)}}{t}dt$.

So $-f(x) = \ln{x} \ln{(1-x)}+f(t)+C = \ln{x} \ln{(1-x)}+f(1-x)+C$. Thus

$f(x)+f(1-x) = C-\ln{x} \ln{(1-x)}$

for some constant $C$. Using our knowledge that $f(0) = 0$, $\displaystyle f(1) = \frac{\pi^2}{6}$, and $\displaystyle \lim_{x \rightarrow 1} \ln{x} \ln{(1-x)}= 0$ by L'Hopital twice, we see that

$f(0)+f(1) = C \Rightarrow C = \frac{\pi^2}{6}$

Then setting $x = \frac{1}{2}$ we obtain $\displaystyle 2f\left(\frac{1}{2}\right) = \frac{\pi^2}{6}-\ln{\frac{1}{2}} \ln{\frac{1}{2}}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 \cdot 2^n} = f\left(\frac{1}{2}\right) = \frac{\pi^2}{12}-\frac{\ln^2{(2)}}{2}$. QED.

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Comment: This was a pretty tough problem that required you to compound a lot of calculus knowledge all into a single problem - series, integration by parts, limits. Recognizing all the steps was the first part; following through with the right computations was another. Still, there weren't really any super clever tricks, mostly just standard substitutions and approaches applied in a somewhat non-standard way. Makes for a very nice problem.

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Practice Problem #1: Show that $\displaystyle \lim_{x \rightarrow 1} \ln{x} \ln{(1-x)} = 0$.

Practice Problem #2: Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n}$. Can you also find $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n^3}$?