# Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$

By using Mathematical Induction, prove the following equation for all positive integers $$n$$: $$\tan^{-1}\frac{1}{2 \cdot 1^{2}} + \tan^{-1}\frac{1}{2 \cdot 2^{2}} +\cdots+\tan^{-1}\frac{1}{2 \cdot n^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$$

I am able to prove the statement for n=1 and I got the assumption statement when $$n = k$$ as follows: $$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times k^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2k+1}$$

The problem is when I have to prove the case for n=k+1. I'm supposed to prove the following: $$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times (k+1)^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2k+3}$$

I simplified the left hand side - with the assumption - into $$\frac{\pi}{4}-\left(\tan^{-1}\frac{1}{2k+1}-\tan^{-1}\frac{1}{2(k+1)^{2}}\right)$$ I understand that I am supposed to use the identity $$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$$ However, I still am unable to obtain the right-hand side from the expression I have right now.

Is there something I am missing?

I don't know what you did wrong but you should get the required result using the given identity: \begin{align} \arctan{\left(\frac{1}{2k+1}\right)}-\arctan{\left(\frac{1}{2(k+1)^2}\right)} &=\arctan{\left(\frac{\frac{1}{2k+1}-\frac{1}{2(k+1)^2}}{1+\left(\frac{1}{2k+1}\right)\left(\frac{1}{2(k+1)^2}\right)}\right)}\\ &=\arctan{\left(\frac{2(k+1)^2-(2k+1)}{(2k+1)(2(k+1)^2)+1}\right)}\\ &=\arctan{\left(\frac{2(k^2+2k+1)-(2k+1)}{2(2k+1)(k^2+2k+1)+1}\right)}\\ &=\arctan{\left(\frac{2k^2+4k+2-2k-1}{2(2k^3+4k^2+2k+k^2+2k+1)+1}\right)}\\ &=\arctan{\left(\frac{2k^2+2k+1}{2(2k^3+5k^2+4k+1)+1}\right)}\\ &=\arctan{\left(\frac{2k^2+2k+1}{4k^3+10k^2+8k+3}\right)}\\ &=\arctan{\left(\frac{2k^2+2k+1}{(2k+3)(2k^2+2k+1)}\right)}\\ &=\arctan{\left(\frac{1}{2k+3}\right)}\\ \end{align}
All you need to do is to prove $$\tan^{-1}\dfrac{1}{2k+1}-\tan^{-1}\dfrac{1}{2(k+1)^{2}} = \tan^{-1} \dfrac{1}{2k+3}.$$ You can use the formula you mentioned in this way: \begin{align}\tan^{-1}\dfrac{1}{2k+1} - \tan^{-1} \dfrac{1}{2k+3} &= \tan^{-1} \dfrac{\frac{1}{2k+1}-\frac{1}{2k+3}}{1+\frac{1}{2k+1}\frac{1}{2k+3}}\\ &= \tan^{-1} \dfrac{2}{(2k+1)(2k+3)+1}\\ &= \tan^{-1} \dfrac{1}{2(k+1)^2}. \end{align}