Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2

[lfdahl]

Show that

\[\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ),\;\;\;\;\; k \geq 1,\]

- and deduce that

\[ \sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}.\]

 

[HallsofIvy]

Looks like an obvious candidate for induction on ki. When k= 1, both sided are $$tan^{-1}(1)$$.

Assume that, for k= j, $$tan^{-1}(j)= \sum_{n=0}^{k-1} tan^{-1}\left(\frac{1}{n^2+ n+ 1}\right)$$. Use that to show that

$$tan^{-1}(j+1)= \sum_{n=0}^{j} tan^{-1}\left(\frac{1}{n^2+ n+ 1}\right)$$.​

 

[kaliprasad]

My solution

Spoiler

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$
Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$
Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$
We get $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)- \tan ^{-1}n$
Adding from 0 to k-1 we get as telescopic sum
Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}k- \tan ^{-1}0 = \tan ^{-1}k$
Taking limit as $k = \infty$
$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}\infty= \frac{\pi}{2}$

 

[lfdahl]

My solution

Spoiler

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$
Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$
Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$
We get $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)- \tan ^{-1}n$
Adding from 0 to k-1 we get as telescopic sum
Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}k- \tan ^{-1}0 = \tan ^{-1}k$
Taking limit as $k = \infty$
$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}\infty= \frac{\pi}{2}$

You´re truly a master, kaliprasad, thankyou very much for your nice solution!(Yes)