Question

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

Answer 1

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:

1. Take `0` and store into `X`. This becomes the current sum.
2. 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.
3. At each iteration, add on `1/N^P` from the previous iteration, then store the result in `X`, the current sum.
4. Display `X`.

So, this would work for finite `p`-series of any length!

Answer 2

`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`