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## IMC 2010 Day 1 Problem 2

Compute the sum of the series
$\displaystyle \sum_{k \geq 0} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$.
IMC 2010 day 1 problem 2

Solution: The answer is not very nice $\displaystyle \frac{\ln 2}{4}-\frac{\pi}{24}$. To get here, split the terms of the series into alternate series, and use integration to calculate the values. To be continued.