Accuracy in Series Approximations
Miller - May 2014

Log[2] = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ...
Can simplify to ∑  1/((2n+1)(2n+2))

a = SetAccuracy[Log[2.], 12]
b = SetAccuracy[Sum[1/((2n+1)(2n+2)),{n,0,500000}],12]
a-b

Pi/4 versus 1 - 1/3 + 1/5 - 1/7 + 1/9 - ..., which simplies to ∑ 2/((4n+1)(4n+3))

x = SetAccuracy[Pi/4,12]
y = SetAccuracy[Sum[2/((4n+1)(4n+3)), {n,0,500000}], 12]
x - y

numdo = 100;
accuracy = 200;
q = SetAccuracy[Pi,accuracy]
r= SetAccuracy[Sum[(4/(8n+1) - 2/(8n+4) - 1/(8n+5) - 1/(8n+6))/16^n, {n,0,numdo}],accuracy]
q - r

prob[days_,n_] := 1.0 - Product[days+1.-k,{k,1,Floor[n]}] / days^Floor[n]

Plot[{.5,prob[365,x]},{x,1,100}]

Plot[{.5,prob[90520,x]},{x,1,1000}]

.5 + Sqrt[2 365 Log[2]]

prob[365,22]
prob[365,23]

prob[365,20]

prob[365,40]