Chapter 2 Problem 5

Evaluate the sum

S=119 1

25 1

49

First lets rewrite this to be

S=1 132 1

52 1

72

Now this looks similar to the Riemann Zeta of 2 just missing some terms so lets rewrite it

S=2n=1

12n2

=214n=1

1n2

But the definition of the Riemann Zeta function is

n=i=1

1in

So our expression for S becomes

S=2142=3

42=

2

8