Question

Find the sum of the series :-

$\rm {tan}^{ - 1} \bigg( \dfrac{2.1}{2 + {1}^{2} + {1}^{4} } \bigg) +{tan}^{ - 1} \bigg( \dfrac{2.2}{2 + {2}^{2} + {2}^{4} } \bigg) +{tan}^{ - 1} \bigg( \dfrac{2.3}{2 + {3}^{2} + {3}^{4} } \bigg) + ... \infty$
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Answer 1

Topic :-

Inverse Trigonometric Function

To Find :-

$\sf {\tan^{-1}\left( \dfrac{2\cdot 1}{2+1^2+1^4}\right)+\tan^{-1}\left( \dfrac{2\cdot 2}{2+2^2+2^4}\right)+\tan^{-1}\left( \dfrac{2\cdot 3}{2+3^2+3^4}\right)\cdots \cdots\infty}$

Solution :-

$\sf {\tan^{-1}\left( \dfrac{2\cdot 1}{2+1^2+1^4}\right)+\tan^{-1}\left( \dfrac{2\cdot 2}{2+2^2+2^4}\right)+\tan^{-1}\left( \dfrac{2\cdot 3}{2+3^2+3^4}\right)\cdots \cdots\infty}$

We can write it as,

$\sf {\displaystyle \sum_{n=1}^{\infty}\tan^{-1}\left( \dfrac{2\cdot n}{2+n^2+n^4}\right)}$

$\sf {\displaystyle \sum_{n=1}^{\infty}\tan^{-1}\left( \dfrac{2\cdot n}{1+(1+n^2+n^4)}\right)}$

$\sf {\displaystyle \sum_{n=1}^{\infty}\tan^{-1}\left( \dfrac{2\cdot n}{1+(n^2+n+1)(n^2-n+1)}\right)}$

$\sf {(\because (n^4+n^2+1)=(n^2+n+1)(n^2-n+1))}$

$\sf {\displaystyle \sum_{n=1}^{\infty}\tan^{-1}\left( \dfrac{(n^2+n+1)-(n^2-n+1)}{1+(n^2+n+1)(n^2-n+1)}\right)}$

$\sf {(\because 2n=(n^2+n+1)-(n^2-n+1))}$

$\sf {\displaystyle \sum_{n=1}^{\infty}\tan^{-1}(n^2+n+1)-\tan^{-1}(n^2-n+1)}$

$\sf{\left(\because \tan^{-1}\dfrac{x-y}{1+xy}=\tan^{-1}x-\tan^{-1}y,where\: x>0\:and\:y>0\right)}$

Now, we will put value of 'n' one by one,

$\sf {\cancel{\tan^{-1}(1^2+1+1)}-\tan^{-1}(1^2-1+1)+}$

$\sf {\cancel{\tan^{-1}(2^2+2+1)}-\cancel{\tan^{-1}(2^2-2+1)}+}$

$\sf {\cancel{\tan^{-1}(3^2+3+1)}-\cancel{\tan^{-1}(3^2-3+1)}+}$

$\cdot$

$\cdot$

$\sf {\displaystyle \lim_{n\to \infty}\tan^{-1}(n^2+n+1)-\cancel{\tan^{-1}(n^2-n+1)}}$

Now, it equals to,

$\sf {\displaystyle \lim_{n\to\infty}\tan^{-1}(n^2+n+1)-\tan^{-1}(1^2-1+1)}$

$\sf {\displaystyle \lim_{n\to\infty}\tan^{-1}(n^2+n+1)-\tan^{-1}(1)}$

$\sf {\tan^{-1}((\infty)^2+\infty+1)-\dfrac{\pi}{4}}$

$\left(\because \tan^{-1}1=\dfrac{\pi}{4}\right)$

$\sf {\tan^{-1}(\infty)-\dfrac{\pi}{4}}$

$\sf {\dfrac{\pi}{2}-\dfrac{\pi}{4}}$

$\left(\because \tan^{-1}\infty=\dfrac{\pi}{2}\right)$

$\sf {\dfrac{\pi}{4}}$

Answer :-

$\sf {\tan^{-1}\left( \dfrac{2\cdot 1}{2+1^2+1^4}\right)+\tan^{-1}\left( \dfrac{2\cdot 2}{2+2^2+2^4}\right)+\cdots\cdots\infty=\dfrac{\pi}{4}}$

Answer 2

Answer:

tan−1(n2+n+1)−4π

m=1∑ntan−1(m4+m2+22m) is equal to

In General formula for tan−1x=tan−1y is

tan−1(1±xyx±y)...(1)

Now, given summation is,

m=1∑ntan−1(m4+m2+22m)=m=1∑ntan−1(1+(m4+m2+1)2m)

Now, factorisation of. m4+m2+1 is

m4+m2+1=(m2+m+1)×(m2−m+1)

and difference between (m2−m+1) and

(m2−m+1) is 2m.

⇒m=1∑ntan−1(1+(m2+m+1)(m2−m+1)(m2+m+1)−(m2−m+1))...(2)

Now, compare eqn(1) & eqn(2)

=m−1∑ntan−1(m2+m+1)−tan−1(m2∗m+1)

Now putting value from 1, 2, ...,n.

=tan−1(m2+1+1)−tan−1(12−1+1)+tan−1(22+2+1)

−tan−1(22−2+1)+...+tan2−1(n2+n+1)..tan−1(n2−n+1)

=tan−1(3)−tan−1(1)+tan−1(7)−tan−1(3)+...+tan−1(n2+n+1)−tan−1(n2−n+1)

here alternative terms will canceled and we are left with tan−1(n2+n+1)−tan−1(1)

So, tan−1(n2+n+1)−4π