Question

If \[a = \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 5 }} + .........\] and \[b = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 4 }} + \dfrac{1}{{\sqrt 6 }} + .........\], then find the value of $\dfrac{{a - b}}{{a + b}}$.
A) $1 - \sqrt 2 $
B) $1 + \sqrt 2 $
C) $1 - \sqrt 3 $
D) $1 + \sqrt 3 $

Answer 1

Hint: We will rewrite b by modifying their terms a bit. We will then get b in terms of a + b and then when we will consider $\dfrac{{a - b}}{{a + b}}$, we will make some modifications and use the b in terms of a +b.

Complete step-by-step answer:
Let us first consider \[a = \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 5 }} + .........\] and \[b = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 4 }} + \dfrac{1}{{\sqrt 6 }} + .........\].
Now, adding both of these, we will get:-
\[ \Rightarrow a + b = \left( {\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 5 }} + .........} \right) + \left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 4 }} + \dfrac{1}{{\sqrt 6 }} + .........} \right)\].
Rewriting it to get as follows:-
\[ \Rightarrow a + b = \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 4 }} + \dfrac{1}{{\sqrt 5 }} + \dfrac{1}{{\sqrt 6 }} + .................\] ………….(1)
Now, consider \[b = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 4 }} + \dfrac{1}{{\sqrt 6 }} + .........\]
We can rewrite it as follows:-
\[ \Rightarrow b = \dfrac{1}{{\sqrt {2 \times 1} }} + \dfrac{1}{{\sqrt {2 \times 2} }} + \dfrac{1}{{\sqrt {2 \times 3} }} + .........\]
Taking $\dfrac{1}{{\sqrt 2 }}$ common, we will get:-
\[ \Rightarrow b = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }} + .........} \right)\]
Looking at the above expression of b and (1), we notice that:-
\[ \Rightarrow b = \dfrac{1}{{\sqrt 2 }}(a + b)\] …………….(2)
Now, considering the expression that we needed to find the values of. It is $\dfrac{{a - b}}{{a + b}}$.
We can rewrite this as $\dfrac{{a - b}}{{a + b}} = \dfrac{{a + b - 2b}}{{a + b}}$
Diving the RHS as follows:-
$ \Rightarrow \dfrac{{a - b}}{{a + b}} = \dfrac{{a + b}}{{a + b}} - \dfrac{{2b}}{{a + b}}$
Simplifying the RHS, we will get as follows:-
$ \Rightarrow \dfrac{{a - b}}{{a + b}} = 1 - \dfrac{{2b}}{{a + b}}$
Now, using (2) in this expression, we will get:-
$ \Rightarrow \dfrac{{a - b}}{{a + b}} = 1 - \dfrac{{2 \times \dfrac{1}{{\sqrt 2 }}(a + b)}}{{a + b}}$
On simplifying the RHS, we will get:
$ \Rightarrow \dfrac{{a - b}}{{a + b}} = 1 - 2 \times \dfrac{1}{{\sqrt 2 }}$
On simplifying the RHS further, we will get:
$ \Rightarrow \dfrac{{a - b}}{{a + b}} = 1 - \sqrt 2 $

Hence, the correct option is (A).

Note: The students must note that they may try to individually find the value of a and b and then put them in the required expression but that could not be solved ever. But you must note that $\dfrac{1}{{\sqrt n }}$ is a divergent series and it can never have a finite sum.
A divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.
The students must remember that $\dfrac{1}{{{n^p}}}$ is a convergent series if p > 1 and divergent if $p \leqslant 1$.
This is not true in case of sequences.
The students must note that it is definitely necessary to modify the question even if the series is given to be convergent because it is never that easy to find the sum of series. As we have seen how easy the question became after modifying it.