How do you find the infinite sum of a p series? EX: #1/n^2#

Question

Emily Ammerman · Accepted Answer

I ended up just making a program to do it in my TI-83 calculator. If anyone wants it, here it is: #"ClrHome"#
#"DelVar X"#
#"DelVar M"#
#"DelVar N"#
#"DelVar P"#
#"Menu(“WHICH SERIES?”, “P-SERIES”, 1, “EXIT”, 2)"#
#"Pause"#
#" "#
#"Lbl 1"#
#"Input “POWER=”, P"#
#"Input “NUM OF TERMS=”, M"#
#0->"X"#
#"For(N, 1, M)"#
#"(X+(1/(N^P)))"->"X"#
#"Disp X"#
#"End"#
#" "#
#"Disp “DONE!”"#
#"Pause"#
#" "#
#"DelVar X"#
#"DelVar M"#
#"DelVar N"#
#"DelVar P"#
#" "#
#"Lbl 2"#
#"ClrHome"#
#"Stop"# Basically: So, this would work for finite #p#-series of any length!

Charles Anctil · Answer

# sum_(n=1)^oo 1/n^s = zeta(s) # where #zeta(s)# is the Riemann zeta function . The complete answer to this question is considerably more complex than it may first appear. I will outline the very basics, and further in-depth analysis is certainly at MSc/PhD level mathematics. The very basic and quick answer is that the infinite sum of a p-series is given by the Riemann zeta function . Where: # zeta(s) = sum_(n=1)^oo 1/n^s # Some specific solutions are: # zeta(2) = sum_(n=1)^oo 1/n^2 = pi^2/6 # (the Bessel Problem ) # zeta(3) = sum_(n=1)^oo 1/n^3 = 1.20205 ... # (Apéry's constant ) # zeta(4) = sum_(n=1)^oo 1/n^4 = pi^4/90 #

Cameron Alexander · Answer

undefined To find the sum of an infinite series of the form 1/n^p, where p is a positive integer, you can use the formula for the sum of a p-series:

Sum = 1/(1^p) + 1/(2^p) + 1/(3^p) + ...

The sum of this series converges if p is greater than 1. The formula to find the sum of a convergent p-series is:

Sum = 1/(1^p) + 1/(2^p) + 1/(3^p) + ... = Σ(1/n^p) = (1/(1^p))/(1 - 1/(2^p))

Where Σ represents the sum, and n starts from 1 and goes to infinity.

For example, if p = 2, the sum of the series 1/n^2 is:

Sum = 1/(1^2) + 1/(2^2) + 1/(3^2) + ... = Σ(1/n^2) = (1/(1^2))/(1 - 1/(2^2)) = 1/(1 - 1/4) = 1/(3/4) = 4/3