Chapter 8.1, Problem 146E

To determine

To write: the first five terms of the sequence and an expression for nth partial sum.

Answer to Problem 146E

The first five terms are

  $\frac{2}{3},\frac{1}{4},\frac{2}{15},\frac{1}{12},\frac{2}{35}$

The expression for nth partial sum is $1-\frac{1}{n+2}$ .

Explanation of Solution

Given information:

Given sequence

  $a_n=\frac{1}{n}-\frac{1}{n+2}$

Calculation:

Let, $a_i$ is the $i^{th}$ term of the sequence then,

The first five terms given by the sequence $a_n=\frac{1}{n}-\frac{1}{n+2}$ are,

  $\begin{array}{l}{a}_1=\frac{1}{1}-\frac{1}{1+2}=1-\frac{1}{3}=\frac{2}{3}\\ a_2=\frac{1}{2}-\frac{1}{2+2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\\ a_3=\frac{1}{3}-\frac{1}{3+2}=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\\ a_4=\frac{1}{4}-\frac{1}{4+2}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\\ a_5=\frac{1}{5}-\frac{1}{5+2}=\frac{1}{5}-\frac{1}{7}=\frac{2}{35}\end{array}$

Hence, the first five terms are

  $\frac{2}{3},\frac{1}{4},\frac{2}{15},\frac{1}{12},\frac{2}{35}$

Now, add all terms up to nth term for expression for nth partial sum

  $\begin{array}{c}{\displaystyle \sum a_n}=\frac{2}{3}+\frac{1}{4}+\frac{2}{15}+\frac{1}{12}+\frac{2}{35}+\mathrm{...}+a_n\\ =1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\mathrm{...}+\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+2}\\ =1-\frac{1}{n+2}\end{array}$

Hence, expression for nth partial sum is $1-\frac{1}{n+2}$ .