Ballantine's Series

Author

Eric W. Weisstein
August 13, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Constants/BallantinesSeries.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/BallantinesSeries.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Series

864 (∑_(n=0)^∞)(((4^n) (((n!))^2))/((325^(n+1)) (2 n+1)!))+1824 (∑_(n=0)^∞)(((4^n) (((n!))^2))/((3250^(n+1)) (2 n+1)!))-20 (cot^(-1))(239)==π

V6

s=864Sum[(((((n!))^2)(4^n))/((2n+1)!(325^(n+1)))),{n,0,∞}]+1824Sum[(((((n!))^2)(4^n))/((2n+1)!(3250^(n+1)))),{n,0,∞}]-20ArcCot[239]

FullSimplify[s]

Log[FullSimplify/@TrigToExp[Exp[s]]]

N[s,20]

V7

$Version

{Sum[(((((n!))^2)(4^n))/((2n+1)!(325^(n+1)))),{n,0,∞}],1824Sum[(((((n!))^2)(4^n))/((2n+1)!(3250^(n+1)))),{n,0,∞}]}//FullSimplify

864Sum[(((((n!))^2)(4^n))/((2n+1)!(325^(n+1)))),{n,0,∞}]+1824Sum[(((((n!))^2)(4^n))/((2n+1)!(3250^(n+1)))),{n,0,∞}]-20ArcCot[239]

FullSimplify[%]

Sum[864(((((n!))^2)(4^n))/((2n+1)!(325^(n+1))))+1824(((((n!))^2)(4^n))/((2n+1)!(3250^(n+1)))),{n,0,∞}]

FullSimplify[%]

FullSimplify[%-20ArcCot[239]]