Question

How do you find the $n^{th}$ partial sum, determine whether the series converges and find the sum when it exists given $\dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ......\dfrac{1}{{n(n + 2)}} + ...$ ?

Answer 1

Hint : First we will start by decomposing the sequence into partial fractions. Then we will substitute values of $n$ in order to evaluate the values of $A,B$ . Then apply the proper limits and evaluate the value of limits.

Complete step-by-step answer :
We will first start by performing a decomposition into partial fractions.
$\dfrac{1}{{n(n + 2)}} = \dfrac{A}{n} + \dfrac{B}{{n + 2}} \\\ = \dfrac{{A(n + 2) + Bn}}{{n(n + 2)}} \;$
Now if we compare the numerators, we get,
$1 = A(n + 2) + Bn$
Now here we use the try and error method in order to evaluate the values of $A,B$ .
In this method we will substitute different values of $n$ in order to evaluate the values of $A,B$ .
So, now when $n = 0$ , the equation becomes,
$1 = 2A \\\ A = \dfrac{1}{2} \;$
So, now when $n = - 2$ , the equation becomes,
$\,\,1 = - 2B \\\ B = - \dfrac{1}{2} \;$
So, the values of $A,B$ are $\dfrac{1}{2}, - \dfrac{1}{2}$ .
Hence,
$\dfrac{1}{{n(n + 2)}} = \dfrac{1}{{2n}} - \dfrac{1}{{2(n + 2)}}$
When $n = 1$
$\dfrac{1}{{1.3}} = \dfrac{1}{2} - \dfrac{1}{6}$
When $n = 2$
$\dfrac{1}{{2.4}} = \dfrac{1}{4} - \dfrac{1}{8}$
When $n = 3$
$\dfrac{1}{{3.5}} = \dfrac{1}{6} - \dfrac{1}{{10}}$
When $n = 4$
$\dfrac{1}{{4.6}} = \dfrac{1}{8} - \dfrac{1}{{12}}$
When $n = 5$
$\dfrac{1}{{5.7}} = \dfrac{1}{{10}} - \dfrac{1}{{14}}$
When $n = n - 2$
$\dfrac{1}{{(n - 2).n}} = \dfrac{1}{{2(n - 2)}} - \dfrac{1}{{2n}}$
When $n = n - 1$
$\dfrac{1}{{(n - 1).(n + 1)}} = \dfrac{1}{{2(n - 1)}} - \dfrac{1}{{2(n + 1)}}$
When $n = n$
$\dfrac{1}{{n.(n + 2)}} = \dfrac{1}{{2n}} - \dfrac{1}{{2(n + 2)}}$
Now, the partial sum is given by,

$${S_n} = \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{{2(n + 1)}} - \dfrac{1}{{2(n + 2)}} \\\ {S_n} = \dfrac{3}{4} - \dfrac{{2n + 3}}{{2(n + 1)(n + 2)}} \;$$

Now we apply the limit here, in order to find where the series converges.
$\mathop {\lim }\limits_{n \to + \infty } \,{S_n} = \dfrac{3}{4}$
Hence, the series converges to $\dfrac{3}{4}$ .
So, the correct answer is “ $\dfrac{3}{4}$ ”.

Note : A limit is a value that a function or a sequence approaches as the input approaches some value. Limits are essential to calculus and mathematical analysis, and are used to define continuity, derivatives and integrals. The concept of limit of a sequence is further generalised to the concept of a limit of a topological net, and is closely related to limit and direct limit in category theory.
Always be sure that all of the terms are of the same type while equating them to zero. While evaluating the value of any term pay close attention to the signs while interchanging the terms. While applying the limits apply proper rules.