Question

Find the sum S:
$\displaystyle \large{S = \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 5} + .... + \frac{1}{n(n+2)}}$
Please show your work too. Thank you!​

Answer 1

Write the sum as

$\displaystyle S = \sum_{k=1}^n \frac1{k(k+2)}$

Decompose the summand into partial fractions; that is, we look for constants a and b such that

$\dfrac 1{k(k+2)} = \dfrac ak + \dfrac b{k+2}$

Condense the right side into a single fraction and set the numerators on both sides equal to one another:

$\dfrac1{k(k+2)} = \dfrac{a(k+2)}{k(k+2)} + \dfrac{bk}{k(k+2)} = \dfrac{a(k+2)+bk}{k(k+2)} \\\\ \implies 1 = a(k+2) + bk = (a+b)k + 2a$

Then a + b = 0 and 2a = 1, so it follows that a = 1/2 and b = -1/2.

We then see that S is a telescoping sum (intermediate terms cancel and the overall sum collapses into a small number of terms):

$\displaystyle S = \frac12 \sum_{k=1}^n \left(\frac1k - \frac1{k+2}\right) \\\\ S = \frac12 \left(1-\frac13 + \frac12 - \frac14 + \frac13 - \frac15 + \cdots + \frac1{n-2} - \frac1n + \frac1{n-1} - \frac1{n+1} + \frac1n - \frac1{n+2}\right) \\\\ S = \frac12 \left(1 + \frac12 - \frac1{n+1} - \frac1{n+2}\right) \\\\ \boxed{S = \frac{n(3n+5)}{4(n+1)(n+2)}}$