Question

If $\theta ={\sum }_{k=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2k^2}\right)$, find $\tan \theta$

Answer 1

To find: $\tan \theta$

If $\theta ={\sum }_{k=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2k^2}\right)$

Let: $\frac{1}{2k^2}=\frac{2}{4k^2}$

Add and Subtract $1$ in Denominator and simplifying using identity : $a^2-b^2=(a+b)(a-b)$

$\frac{1}{2k^2}=\frac{2}{4k^2}=\frac{(2k+1)-(2k-1)}{1+4k^2-1^2}$

$\frac{1}{2k^2}=\frac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}$

As we know, ${\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(b\right)={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$

Let: $a=(2k+1)$ and $b=(2k-1)$

Thus, we have:

${\sum }_{k=1}^{\infty }{\tan }^{-1}\frac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}={\sum }_1^{\infty }\left[{\tan }^{-1}\right(2k+1)-{\tan }^{-1}(2k-1\left)\right]$

Here we observe: $2k+1$ is $(k+1{)}^{th}$ term of $2k-1$

So, by using "Method of Difference"

i.e $(a_3-a_1)+(a_5-a_3)+(a_7-a_5)+...+(a_{2n+1}-a_{2n-1})+.......$

Noting that all terms except $a_1$ and $a_{2n+1}$ will cancel

As when $k$ tends to $\infty$ , $2k+1$ tends to $\infty$ and $a_1$ remains, we get

${\sum }_{k=1}^{\infty }\left[{\tan }^{-1}\right(2k+1)-{\tan }^{-1}(2k-1\left)\right]=f\left(\infty \right)-f\left(1\right)$

Since:

${\tan }^{-1}\left(1\right)=\frac{\pi }{4}$ and: $\underset{k\to \infty }{lim}{\tan }^{-1}(2k+1)=\frac{\pi }{2}$

We finally obtain:

${\sum }_{k=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2k^2}\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

Thus, $\theta =\frac{\pi }{4}$

$\tan \left(\frac{\pi }{4}\right)=1$