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

##### 2 Answers

I ended up just making a program to do it in my TI-83 calculator. If anyone wants it, here it is:

Basically:

- Take
#0# and store into#X# . This becomes the current sum. - For the variable
#N# (the current term in the series#sum_(N=1)^(M) 1/(N^P)# ), we go through iterations from the#1# st iteration to the#M# th iteration, where#M# is the number of terms. - At each iteration, add on
#1/N^P# from the previous iteration, then store the result in#X# , the current sum. - Display
#X# .

So, this would work for finite

# sum_(n=1)^oo 1/n^s = zeta(s) # where#zeta(s)# is the Riemann zeta function .

#### Explanation:

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 #