Chapter 8, Problem 12CR

To determine

The sum of the given series.

Answer to Problem 12CR

The sum of the given series is $\frac{15551}{2520}$ .

Explanation of Solution

Given: The given series is represented as-

  ${\displaystyle \sum _{i=1}^8\frac{i}{i+1}}$

Concepts Used: The sum of any series can be represented using the summation sign as follows-

  $a_1+a_2+a_3+\mathrm{....}+a_n={\displaystyle \sum _{k=1}^n{a}_k}$

This represents a series whose $n^{th}$ term is $a_k$ and the sequence consists of n terms which are $a_1,a_2,a_3,\mathrm{....},a_n$ , i.e.- putting in k = 1, 2, 3 …. in the $n^{th}$ term of the sequence gives the respective terms of the series. Also,

  ${\displaystyle \sum _{k=1}^n(a_k\pm b_k)}={\displaystyle \sum _{k=1}^n{a}_k}\pm {\displaystyle \sum _{k=1}^n{b}_k}$

  ${\displaystyle \sum _{k=1}^{n}c}=c{\displaystyle \sum _{k=1}^{n}1}=cn$ , for any real constant c.

Calculations: The given series can be expanded as follows-

  ${\displaystyle \sum _{i=1}^8\frac{i}{i+1}}$

  $={\displaystyle \sum _{i=1}^8\frac{i+1-1}{i+1}}$

  $={\displaystyle \sum _{i=1}^8\left(\frac{i+1}{i+1}-\frac{1}{i+1}\right)}$

  $={\displaystyle \sum _{i=1}^8\left(1-\frac{1}{i+1}\right)}$

  $={\displaystyle \sum _{i=1}^{8}1}-{\displaystyle \sum _{i=1}^8\frac{1}{i+1}}$

  $=8-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)$

  $=8-\left(\frac{4609}{2520}\right)$

  $=\frac{20160-4609}{2520}$

  $=\frac{15551}{2520}$

Thus the sum of the given series is

  $\frac{15551}{2520}$ .