Transcript
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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