<s>[INST] Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$ Find the value of $$\sum^{n=k}_{n=1}\tan^{-1}\frac{4}{4n^2+3}$$ 
I tried multiplying numerator and denominator by $n^2$, but got nothing. How do I split the term inside $\tan^{-1}$?
 [/INST] I tried again and got the answer. $\frac{4}{4n^2 + 3} = \frac{1}{n^2 + 3/4} = \frac{1}{1 + n^2 -1/4} = \frac{1}{1 + (n+1/2)(n-1/2)} = = \frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)}$  
$$\sum^{n=k}_{n=1}\tan^{-1}\frac{4}{4n^2+3} = \sum^{n=k}_{n=1}\tan^{-1}\frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)} = \boxed{\tan^{-1}(k+1/2) - \tan^{-1}1/2}$$ 
</s>