<s>[INST] 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?
 [/INST] I don't know what you did wrong but you should get the required result using the given identity:
$$\begin{align}
\arctan{\left(\frac{1}{2k+1}\right)}-\arctan{\left(\frac{1}{2(k+1)^2}\right)}
&=\arctan{\left(\frac{\frac{1}{2k+1}-\frac{1}{2(k+1)^2}}{1+\left(\frac{1}{2k+1}\right)\left(\frac{1}{2(k+1)^2}\right)}\right)}\\
&=\arctan{\left(\frac{2(k+1)^2-(2k+1)}{(2k+1)(2(k+1)^2)+1}\right)}\\
&=\arctan{\left(\frac{2(k^2+2k+1)-(2k+1)}{2(2k+1)(k^2+2k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+4k+2-2k-1}{2(2k^3+4k^2+2k+k^2+2k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{2(2k^3+5k^2+4k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{4k^3+10k^2+8k+3}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{(2k+3)(2k^2+2k+1)}\right)}\\
&=\arctan{\left(\frac{1}{2k+3}\right)}\\
\end{align}$$
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