Question

The sum up to infinity terms of the series $\dfrac{4}{5} + \dfrac{8}{{65}} + \dfrac{{12}}{{325}} + ... + \dfrac{{ar}}{{b{r^4} + c}} + ...$ is (Here, $a,b,c$ are integers and $\dfrac{{ar}}{{b{r^4} + c}}$ denotes the ${r^{th}}$ term of the series).
A) $\dfrac{1}{2}$
B) $\dfrac{1}{3}$
C) $\dfrac{1}{4}$
D) $1$

Answer 1

Hint:
We are given the general formula of ${r^{th}}$ term. Substituting different values for $r$ and comparing with the given terms we get the value of $a, b$ and $c$. Then we can consider the partial sum of the expression and see the infinite sum by tending $n$ to infinity.

Useful formula:
For any $a,b$ we have the algebraic identities.
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${a^2} - {b^2} = (a - b)(a + b)$

Complete step by step solution:
It is given that the \[{r^{th}}\] term of the series is $\dfrac{{ar}}{{b{r^4} + c}}$
So substituting $r = 1$, we get the first term.
That is, $\dfrac{{a \times 1}}{{b \times {1^4} + c}} = \dfrac{a}{{b + c}}$
But we have, first term given as $\dfrac{4}{5}$.
This gives, $\dfrac{a}{{b + c}} = \dfrac{4}{5}$.
So we have, $a = 4,b + c = 5$
Now substitute $r = 2$,
$\dfrac{{a \times 2}}{{b \times {2^4} + c}} = \dfrac{{2a}}{{16b + c}}$
But we have, second term as $\dfrac{8}{{65}}$.
Comparing we get,
$16b + c = 65$
Also $b + c = 5$
Subtracting both sides we get,
$15b = 60$
Dividing both sides by $15$ we get,
$b = \dfrac{{60}}{{15}} = 4$
$ \Rightarrow c = 5 - b = 5 - 4 = 1$
So we have, $a = 4,b = 4,c = 1$
Therefore the ${r^{th}}$ term becomes $\dfrac{{4r}}{{4{r^4} + 1}}$.
Consider $4{r^4} + 1$.
Adding and subtracting $4{r^2}$ in the right side we get,
$4{r^4} + 1 = 4{r^4} + 4{r^2} + 1 - 4{r^2}$
Now we have,
$4{r^4} + 1 = (4{r^4} + 4{r^2} + 1) - 4{r^2}$
We have ${(a + b)^2} = {a^2} + 2ab + {b^2}$
Using this we get,
$4{r^4} + 1 = {(2{r^2} + 1)^2} - 4{r^2}$
We know, ${a^2} - {b^2} = (a - b)(a + b)$
$ \Rightarrow 4{r^4} + 1 = [(2{r^2} + 1) - 2r][(2{r^2} + 1) + 2r]$
$ \Rightarrow 4{r^4} + 1 = (2{r^2} + 1 - 2r)(2{r^2} + 1 + 2r)$

Then $\dfrac{{4r}}{{4{r^4} + 1}} = \dfrac{{4r}}{{(2{r^2} + 1 - 2r)(2{r^2} + 1 + 2r)}}$
$\dfrac{{4r}}{{4{r^4} + 1}} = \dfrac{1}{{(2{r^2} + 1 - 2r)}} - \dfrac{1}{{(2{r^2} + 1 + 2r)}}$
Now consider the partial sum,
$\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = \sum\limits_1^n {(\dfrac{1}{{(2{r^2} + 1 - 2r)}} - \dfrac{1}{{(2{r^2} + 1 + 2r)}}} )$
Substituting values $r = 1,2,3...$ we get,
\[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = \dfrac{1}{{2 + 1 - 2}} - \dfrac{1}{{2 + 1 + 2}} + \dfrac{1}{{2 + 1 + 2}} - \dfrac{1}{{8 + 1 + 4}} + \dfrac{1}{{8 + 1 + 4}} - ...\]
\[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = 1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{{13}} + \dfrac{1}{{13}} - ...\]
So we can see that the consecutive terms get cancelling.
When $n \to \infty $, \[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = 1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{{13}} + \dfrac{1}{{13}} - ... \to 1\]
So we get the sum up to infinity of the series is $1$.

$\therefore $ The answer is option D.

Note:
This problem includes a lot of steps. So we may have made mistakes easily. Also remember that when $n \to \infty $, then $\dfrac{1}{n} \to 0$ and so $1 - \dfrac{1}{n} \to 1$. We often use the idea of partial sum in finding the infinite sum.