Friday, June 29, 2007

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} $.



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.


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


  1. ln tan x = ln sin x - ln cos x = ln sin x - ln sin (pi/2 - x)

  2. how come u only post stuff about calculus now??

    what happened to ... funner math?

  3. o_O. I would think calculus would be easier for most people to understand for one thing... and other problems are harder to solve.

  4. oh. fine. well. other than calc

    we are going to the zoo on friday cuz liz cooledge is back. i have the details on ur face book, but i think you check this more often....

    hope you can come!