Question

Use the identity$\frac{1}{\left(\mathrm{k}\right(\mathrm{k}+1\left)\right)}=\frac{\frac{1}{\mathrm{k}-1}}{(\mathrm{k}+1)}$ and Exercise 35 to compute$\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}$

Short answer

Hence, we conclude that $\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}=1-\frac{1}{n+1}$

Step-by-step solution

Step 1:

To compute $\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}$

Given : $\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}=\sum _{k=1}^n \frac{1}{k}-\frac{1}{(k+1)}\text{and}\sum _{i=1}^n \left(a_i-a_{i-1}\right)=\left(a_n-a_0\right)$

Proof : $\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}=-\sum _{k=1}^n \left(\frac{1}{(k+1)}-\frac{1}{k}\right)$

Let $a_k=\frac{1}{(k+1)}$

$-\sum _{k=1}^n \left(\frac{1}{(k+1)}-\frac{1}{k}\right)=-\sum _{k=1}^n \left(a_k-a_{k-1}\right)$

Step 2:

Since, $\sum _{i=1}^n \left(a_i-a_{i-1}\right)=\left(a_n-a_0\right)$

We have that

$-\sum _{k=1}^n \left(a_k-a_{k-1}\right)=-\left(a_n-a_0\right)=\left(a_0-a_n\right)$

Therefore,$\sum _{k=1}^n \frac{1}{\left(k\right(k+1\left)\right)}=a_0-a_n=1-\frac{1}{n+1}$