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?

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:
$$ \begin{align}\tan^{-1}\dfrac{1}{2k+1} - \tan^{-1} \dfrac{1}{2k+3} &= \tan^{-1} \dfrac{\frac{1}{2k+1}-\frac{1}{2k+3}}{1+\frac{1}{2k+1}\frac{1}{2k+3}}\\
&= \tan^{-1} \dfrac{2}{(2k+1)(2k+3)+1}\\
&= \tan^{-1} \dfrac{1}{2(k+1)^2}.
\end{align}$$
