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

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

Basically:

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# 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:

Some specific solutions are:

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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

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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