Welcome to NRICH.

 
S¥ n = 1 tan-1[2/n2] = 3p/4
By Arun Iyer on Sunday, November 11, 2001 - 06:52 pm:

could any one please show that......
S¥ n = 1 tan-1[2/n2] = 3p/4

love arun

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

If you write tan-1[2/n2] = arg(1 + 2i/n2) then the left hand side becomes:

arg(1 + 2i/12) + arg(1 + 2i/22) + arg(1 + 2i/32) + ...

= arg[(1 + 2i/12)(1 + 2i/22)(1 + 2i/32)...]

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

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

sin x = x Pn=1¥(1 - x2/(n2p2)).

Then it only remains to take the argument, and hopefully that will work out as 3p/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 limn->¥ tan-1 -n(n+3)/(n+1)(n-2) = tan-1(-1) = 3p/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=-p

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