Question

How do you test the series $\sum{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}}$ from n is $\left[ 0,\infty \right)$ for convergence?

Answer 1

Hint: The first criteria would be to find the limit of the sum if the value of $n\to \infty $. We would break the series in its individual terms and express its difference of two different terms. Then we find the limit value to find the convergence and converging point.

Complete step-by-step solution:
We need to find the convergence of the series $\sum{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}}$ where the summation is for $n\in \left[ 0,\infty \right)$.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. Here ${{t}_{n}}=\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}$.
We try to form the term as the difference of two different terms.
We express the numerator 1 as $1=\left( n+2 \right)-\left( n+1 \right)$.
So, ${{t}_{n}}=\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}=\dfrac{\left( n+2 \right)-\left( n+1 \right)}{\left( n+1 \right)\left( n+2 \right)}=\dfrac{1}{\left( n+1 \right)}-\dfrac{1}{\left( n+2 \right)}$.
We need to find the sum of $\sum\limits_{n=0}^{\infty }{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}}$. We can express it as $\sum\limits_{n=0}^{\infty }{\left[ \dfrac{1}{\left( n+1 \right)}-\dfrac{1}{\left( n+2 \right)} \right]}$.
We try to find the terms of the series and put the values of $n=0,1,2,3,.....$.
For $n=0$, we get ${{t}_{0}}=\dfrac{1}{1}-\dfrac{1}{2}$.
For $n=1$, we get ${{t}_{1}}=\dfrac{1}{2}-\dfrac{1}{3}$.
For $n=2$, we get ${{t}_{2}}=\dfrac{1}{3}-\dfrac{1}{4}$.
The summation is $\sum\limits_{n=0}^{\infty }{{{t}_{n}}}={{t}_{0}}+{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+.....$.
So, $\sum\limits_{n=0}^{\infty }{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}}=\left( \dfrac{1}{1}-\dfrac{1}{2} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+....$
The simplified form becomes
\[\begin{align}
  & \sum\limits_{n=0}^{\infty }{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}} \\
 & =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3}+.... \\
\end{align}\]
The terms get cancelled in a particular way where the last part of a term is cancelled by the first part of its successor.
The end is where we have to put the value of $n=\infty $ in $-\dfrac{1}{\left( n+2 \right)}$.
The limit value of the term is 0.
The summation tends to 0 as the value of n tends to $\infty $.
Therefore, the series $\sum{\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}}$ is converging and it is converging to 1.

Note: The convergence can also be found from the relation of $\dfrac{{{t}_{n+1}}}{{{t}_{n}}}$. If the ratio is less than 1 then we can say the series is converging. In this case the ratio is less than 1 as the value of n is in natural numbers.