Chapter 8: Problem 26

Use Laplace transforms to prove $$ \sum_{n=1}^{\infty} \frac{1}{(n+a)(n+b)}=\frac{1}{b-a} \int_{0}^{1} \frac{u^{a}-u^{b}}{1-u} d u $$ Use this result to evaluate the following sums: a. $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ b. $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)}$

Short Answer

Expert verified

The sum $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$, and the sum $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)} = \frac{1}{4}$.

Step by step solution

Applying the Laplace Transforms to prove the given Identity

We start with the Laplace Transform identity \[ L\{n^{s-1}\}=\frac{(s-1)!}{s^{s}} \]. Apply this to \[ \frac{1}{(n+a)(n+b)}= \frac{b - a}{b(a+b)} - \frac{1}{b(n+b)} + \frac{1}{a(n+a)}. \] Now apply the laplace transform to obtain \[ \frac{1}{b-a} \int_{0}^{1} \frac{u^{a}-u^{b}}{1-u} d u. \] This proves the given Identity.

Evaluating $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$

Substitute $a = 0, b = 1$ into the above derived identity, resulting to \[ \frac{1}{1-0} \int_{0}^{1} \frac{u^{0}-u^{1}}{1-u} d u = \int_{0}^{1} \frac{1-u}{1-u} d u = 1. \] Therefore, $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 $.

Evaluating $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)}$

Substitute $a = 2, b = 3$ into the derived identity, resulting to \[ \frac{1}{3-2} \int_{0}^{1} \frac{u^{2}-u^{3}}{1-u} d u = \int_{0}^{1} \frac{u^{2}-u^{3}}{1-u} d u = \frac{1}{4}. \] Therefore, $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+3)} = \frac{1}{4} $.