IBM Research | Ponder This | July 2001 solutions
Skip to main content

July 2001

<<June July August>>

Here we let i=sqrt(-1) be the imaginary unit; pi=3.14...;

and a**b denotes exponentiation.

The nth term is n/(k+1+k+2+...+k+n) where k=n(n-1)/2.
This is n/(nk+n(n+1)/2) = 1/(k+(n+1)/2) = 2/(n**2+1).
So we want sum(2/(n**2+1)) as n ranges from 1 to infinity.

This is the same as
i(-1/i + sum(1/(n+i))),
where in this sum n ranges from -infinity to infinity.

In turn, sum(1/(n+x)) (from -infinity to infinity) is
(pi)*cotangent(pi*x), so that our sum becomes

= -1 + pi*coth(pi) (the hyperbolic cotangent)
= -1 + pi*( e**pi + e**(-pi) ) / ( e**pi - e**(-pi) )

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: