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

Jun 24, 2017

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

$\text{ClrHome}$
$\text{DelVar X}$
$\text{DelVar M}$
$\text{DelVar N}$
$\text{DelVar P}$
$\text{Menu(“WHICH SERIES?”, “P-SERIES”, 1, “EXIT”, 2)}$
$\text{Pause}$
$\text{ }$
$\text{Lbl 1}$
$\text{Input “POWER=”, P}$
$\text{Input “NUM OF TERMS=”, M}$
$0 \to \text{X}$
$\text{For(N, 1, M)}$
$\text{(X+(1/(N^P)))"->"X}$
$\text{Disp X}$
$\text{End}$
$\text{ }$
$\text{Disp “DONE!”}$
$\text{Pause}$
$\text{ }$
$\text{DelVar X}$
$\text{DelVar M}$
$\text{DelVar N}$
$\text{DelVar P}$
$\text{ }$
$\text{Lbl 2}$
$\text{ClrHome}$
$\text{Stop}$

Basically:

1. Take $0$ and store into $X$. This becomes the current sum.
2. For the variable $N$ (the current term in the series ${\sum}_{N = 1}^{M} \frac{1}{{N}^{P}}$), 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 $\frac{1}{N} ^ P$ 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!

Jun 25, 2017

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ s = \zeta \left(s\right)$ where $\zeta \left(s\right)$ 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:

$\zeta \left(s\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ s$

Some specific solutions are:

$\zeta \left(2\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$ (the Bessel Problem )

$\zeta \left(3\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 3 = 1.20205 \ldots$ (Apéry's constant )

$\zeta \left(4\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 4 = {\pi}^{4} / 90$