Addition At Its Finest. Topic: Calculus/S&S.

Problem: Evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} $ where $ x $ is a real number with $ |x| < 1 $.

Solution: Looking at that all too common denominator, we do a partial fraction decomposition in hopes of telescoping series. The summation becomes

$ \displaystyle \sum_{n=1}^{\infty} \left(\frac{x^n}{n}-\frac{x^n}{n+1}\right) $.

Common Taylor series knowledge tells us that

$ \displaystyle \ln{(1-x)} = -\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right) = -\sum_{n=1}^{\infty} \frac{x^n}{n} $,

which convenient fits the first part of the summation. As for the second part, we get

$ \displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n+1} = \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = \frac{-\ln{(1-x)}-x}{x} $

from the same Taylor series. Combining the results, our answer is then

$ \displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = 1-\ln{(1-x)}+\frac{\ln{(1-x)}}{x} $.

QED.

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Comment: Even though the trick at the beginning didn't actually get much to telescope, the idea certainly made it easier to recognize the Taylor series. Algebraic manipulations are nifty to carry around and can be applied in problems wherever you go.

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Practice Problem: Show that $ \displaystyle \int_0^{\frac{\pi}{2}} \ln{(\tan{x})} = 0 $.