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Partial Fractions

Express rational functions as a sum of fractions with simpler denominators. You can apply this to telescoping series to mass cancel terms in a seemingly complicated sum.

Level 4

         

\[ \frac {1}{1 \times 2} + \frac {1}{1 \times 5} + \frac {1}{3 \times 3} + \frac {1}{2 \times 7} + \frac {1}{5 \times 4} + \ldots \]

Find the sum of this infinite series.

\( S = \frac {1}{1\times 2 \times 3} + \frac {1}{2\times 3 \times 4} + \ldots + \frac {1} {(n)(n+1)(n+2)} \) \(+ \ldots +\frac {1}{14 \times (14+1) \times (14+2)} = \frac {a} {b}, \) where \(a\) and \(b\) are positive, coprime integers. What is the value of \( a + b \)?

\[\displaystyle{\sum _{ n=1 }^{ 2015 }{ \frac { { n }^{ 2 }+n+1 }{ \left( { n }^{ 2 }+n \right) \left( n+1 \right) ! } }}\] can be represented in the form \(\dfrac { a }{ b } \), where \(a\) and \(b\) are coprime positive integers. Find the value of \(a-b\).

\[\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{2015^{2^{n}} - 2015^{-(2^{n})}} \]

If the closed form of the series above is in the form \( \frac a b \), where \(a\) and \(b\) are positive coprime integers, then find \(b - a.\)

For each positive integer \(n\), let \[H_n = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}.\] If \[ \sum_{n=4}^{\infty} \frac{1}{nH_nH_{n-1}} = \frac{a}{b} \] for relatively prime positive integers \(a\) and \(b\), find \(a+b\).

This problem is shared by Sandeep S.

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