If $\sum_{n=1}^\infty\tan^{-1}\left(\frac4{n^2+n+16}\right)=\tan^{-1}\left(\frac\alpha{ 10 }\right)$, then find $\alpha$.

Question

$$\sum _ { n = 1 } ^ { \infty } \tan ^ { -1 } \left( \frac { 4 } { n ^ { 2 } + n + 16 } \right)= \tan ^ { -1 } \left( \frac { \alpha } { 10 } \right)$$ Find $\alpha$.

I know I need to convert to $$\operatorname{arctan}\frac{a-b}{1+ab}$$ ut here I am not able to do so.

Answer 1

Use $$\arctan\frac{4}{r^2+r+16}=\arctan\frac{\frac{r+1}{4}-\frac{r}{4}}{1+\frac{r+1}{4}\cdot\frac{r}{4}}$$

Answer 2

Check out this solution , hope you understand . Ask if you face any problem

Answer 3

Hint:

Let $f(m)=\arctan(am+b),$

$f(n+1)-f(n)$

$=\arctan\dfrac{a(n+1)+b-(an+b)}{1+(an+b)(an+a+b)}$ $=\arctan\dfrac a{a^2n^2+n(2ab+a^2)+ab+b^2+1}$

Compare the coefficients of $n^2,n,n^0$

$$\dfrac a{a^2}=\dfrac41\iff a=\dfrac14$$

$$\dfrac14=\dfrac{2ab+a^2}a=2b+a\iff b=0$$

$$\dfrac4{16}=\dfrac a{ab+b^2+1}=a$$

Try the same for $\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$