2
$\begingroup$

Show that

$$\tan^{-1}\left({1\over nx}\right)=\sum_{i=n}^{\infty}\tan^{-1}\left({1\over i(i+1)x+x^{-1}}\right)\tag1$$

$(n,x)$ $\ge1$; x is any real numbers

My try:

Setting $(x,n)=1$

We have $$\sum_{k=1}^{\infty}\tan^{-1}\left({1\over k^2+k+1}\right)={\pi\over 4}$$

We know that

$$\tan^{-1}\left({1\over a}\right)+\tan^{-1}\left({1\over b}\right)=\tan^{-1}\left({a+b\over ab-1}\right)$$

Well known

$$\tan^{-1}\left({1\over 2}\right)+\tan^{-1}\left({1\over 3}\right)={\pi\over 4}$$

Let $k=1,2$ then

$${a+b\over ab-1}={1\over 3}$$

$${a+b\over ab-1}={1\over 7}$$

When I solve these simultaneous equations I got $a=b$. Sub to the top I got

$a^2-6a-1=0$, complete the square, then $a=3\pm\sqrt{10}$

Anyway I don't think I am on the right track here, please help to prove (1)

  • 3
    The identity itself is suggesting that $$ \arctan \frac{1}{nx} - \arctan\frac{1}{(n+1)x} = \arctan\left(\frac{x}{1+n(n+1)x^2}\right). $$ Can you check this using the addition formula for $\arctan$ you pointed out? Also, can you show that the identity does follow from this? – 2017-01-02

0 Answers 0