Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$

Question

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 $\tan^{-1}$?

Answer 1

The trick is to turn it into a telescoping series. Since

$$\tan (a - b) = \frac{\tan a - \tan b}{1+\tan a \tan b},$$

the thing to try is to write $\frac{4}{4n^2+3}$ in the form $\frac{c_n-c_{n+1}}{1+c_nc_{n+1}}$. A little experimentation leads to

$$c_n = \frac{1}{n-\frac{1}{2}}.$$

Answer 2

I tried again and got the answer. $\frac{4}{4n^2 + 3} = \frac{1}{n^2 + 3/4} = \frac{1}{1 + n^2 -1/4} = \frac{1}{1 + (n+1/2)(n-1/2)} = = \frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)}$

$$\sum^{n=k}_{n=1}\tan^{-1}\frac{4}{4n^2+3} = \sum^{n=k}_{n=1}\tan^{-1}\frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)} = \boxed{\tan^{-1}(k+1/2) - \tan^{-1}1/2}$$