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BITSAT 2018 Slot 2 Solved Paper

Section: Mathematics

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Question : 116 of 150

Marks: +1, -0

The value of

S=\sum\limits_{n=1}^{\infty}\tan^{-1}\frac{2n}{n^{4}+n^{2}+2}

is equal to

$\frac{\pi}{2}$
$\pi$
$\frac{\pi}{4}$
None of these

Solution:

Given,

S=\sum\limits_{n=1}^{\infty}\tan^{-1}\frac{2n}{n^{4}+n^{2}+2}

.......(i)

Let

n^{4}+n^{2}+1=[(n^{2})^{2}+1^{2}

+2(n^{2})(1)]-n^{2}

=(n^{2}+1)^{2}-n^{2}

=(n^{2}+n+1)(n^{2}-n+1)

.........(ii)

Let

u_{n}=\tan^{-1}\left(\frac{2n}{n^{4}+n^{2}+2}\right)

=\tan^{-1}\frac{2n}{1+(n^{4}+n^{2}+1)}

=\tan^{-1}\frac{(n^{2}+n+1)-(n^{2}-n+1)}{1+(n^{2}+n+1)(n^{2}-n+1)}

u_1=\tan^{-1}(n^{2}+n+1)

-\tan^{-1}(n^{2}-n+1)

On putting n = 1, 2 , 3 ....... successively in Eq. (iii), we get

u_{1}=\tan^{-1}3-\tan^{-1}1

u_{2}=\tan^{-1}7-\tan^{-1}3

u_{3}=\tan^{-1}13-\tan^{-1}7

.................................

u_{n}=\tan^{-1}(n^{2}+n+1)-\tan^{-1}

(n^{2}-n+1)

On adding vertically, we get

\sum\limits_{n=1}^{\infty}u_{n}=\tan^{-1}(n^{2}+n+1)-\tan^{-1}1

S=\lim\limits_{n\rightarrow\infty}\sum\limits_{n=1}^{\infty}u_{n}

[from Eq. (i)]

=\lim\limits_{n\rightarrow\infty}\tan^{-1}(n^{2}+n+1)-\tan^{-1}1

=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}

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