**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} $?

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