could any one please show that......

¥ å n=1

tan-1[2/n2]=3p/4

love arun

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) + ¼ = arg[(1 + 2i/1² )(1 + 2i/2² )(1 + 2i/3² )¼]

(using the fact that argx + argy = argx y).

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

sinx = x

¥ Õ n=1

(1 - x2 /(n2 p2 ))

.

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

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.)

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