$\sum_{n = 1}^{\infty} \tan^{- 1} [2 / n^{2}] = 3 π / 4$

By Arun Iyer on Sunday, November 11, 2001 - 06:52 pm:

could any one please show that...... $\sum_{n = 1}^{\infty} \tan^{- 1} [2 / n^{2}] = 3 π / 4$
love arun

By Michael Doré on Sunday, November 11, 2001 - 07:46 pm:

If you write $\tan^{- 1} [2 / n^{2}] = \arg (1 + 2 i / n^{2})$ then the left hand side becomes:

$\arg (1 + 2 i / 1^{2}) + \arg (1 + 2 i / 2^{2}) + \arg (1 + 2 i / 3^{2}) + \dots$ $= \arg [(1 + 2 i / 1^{2}) (1 + 2 i / 2^{2}) (1 + 2 i / 3^{2}) \dots]$

(using the fact that $\arg x + \arg y = \arg x y$ ).

It is then straightforward to evaluate the product in square brackets using Euler's sine product, i.e.

$\sin x = x Π_{n = 1}^{\infty} (1 - x^{2} / (n^{2} π^{2}))$ .

Then it only remains to take the argument, and hopefully that will work out as $3 π / 4$ .

By David Loeffler on Monday, November 12, 2001 - 02:22 pm:

I think there's an alternative way of doing this - I saw it in a book of problems ages ago.* It gave an explicit form for S(n), the sum of the first n terms, as tan^-1 (p(n)/q(n)) where p and q were polynomials.

After calculating the first few S(n) we find that p(n) = n(n+3) and q(n) = (n-2)(n+1); it is easy to check that tan^-1 (p(n+1)/q(n+1)) - tan^-1 (p(n)/q(n)) = (p(n+1)/q(n+1) - p(n)/q(n)) / (1 + p(n)p(n+1)/q(n)q(n+1)) simplifies down to 2/(n+1)² , as required. So having checked that the first few S are right, this follows for all n by induction.

Hence the infinite sum is equal to lim_{n-> infinity} tan^-1 -n(n+3)/(n+1)(n-2) = tan^-1 (-1) = 3pi/4.

(PS. It was "The Red Book of Mathematical Problems", by Williams and Hardy, Dover 1996, problem 26; but I think they got it from someone else. Their method was actually rather different from mine above, as they dealt with odd-numbered terms and even-numbered terms separately.)

By Arun Iyer on Tuesday, November 13, 2001 - 08:18 pm:

Michael and David,
Thanks for these smooth and easy answers!!!

The solution I had before was quite messy.....

1. I started with Euler's sine product( given by Michael above),
2. Then replaced x by a+ib,
3. Then took log of both the sides,
4. Equated imaginary parts on both sides
5. Replaced a=b=-pi

and that gave me the answer........
It's quite tedious and totally messy.....

Believe me!! It is nothing like the answers you both have given!!

thanks again...
love arun