AdrianZ said:Do you know anything about series? It's a series question. and your series shows different behaviors depending what p you use. for example for p=1/2 the series is the harmonic series which is a famous divergent series. for other values you can use theorems or series tests. for example if p<0 then the series is obviously divergent because it fails the necessary condition to converge. I'll give you a hint.
separate values of p into different groups. like 2p<0, 2p=1, 0<2p<1, 2p>1. That'll be the first step to recognize what the problem asks you to do. I gave you the answer for p<0 & p=1. I let you solve the problem for the other 2 cases. you may not even need to consider the case that 0<2p<1.
About calculating its values, you have to know what p is. moreover, It's usually hard and sometimes extremely difficult to calculate the value of a series.
ghc said:p is a positive integer.
The formula for calculating the sum (1/n)^p is:
∑(1/n)^p = 1/1^p + 1/2^p + 1/3^p + ... + 1/n^p.
In order to determine if the sum (1/n)^p converges or diverges, we can use the p-series test. If p > 1, then the series converges. If p ≤ 1, then the series diverges.
The value of p in the sum (1/n)^p determines whether the series converges or diverges. If p > 1, then the series converges, and if p ≤ 1, then the series diverges.
No, the sum (1/n)^p can only converge for values of p that are greater than 1. If p ≤ 1, then the series will diverge.
Yes, the sum (1/n)^p can converge for negative values of p. However, it will only converge if p is a negative even integer. For example, if p = -2, then the series will converge to π^2/6. If p is a negative odd integer, the series will diverge.