Convergence of the Series (1/n)^p: A Proof for Positive Integer p

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In summary: But to be completely rigorous, you may need to show that the series is absolutely convergent for all the cases in which it converges. That is, if you take the absolute value of each term in the series, it will still converge. This can be shown using the comparison test with the geometric series. In summary, the series \sum_{n=2}^{+\infty}(\frac{1}{n})^{2p} where p \in N can be analyzed by separating values of p into different groups and using theorems or series tests to determine convergence. It is also necessary to consider absolute convergence in order to prove the convergence of the series. Calculating the value of the series can be difficult and sometimes impossible
  • #1
ghc
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[itex]\sum_{n=2}^{+\infty}(\frac{1}{n})^{2p}[/itex] where [itex]p \in N[/itex].
Does this sum converge? How do we calcultate it?

I am not very familiar with maths, so if there is some theorem that can used please tell it to me.
Thank you.
 
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  • #2
We don't do your homework for you at this site, and this looks very much like homework. (And even if it isn't, we treat questions that look very much like homework as if they were homework.)

Try showing some work. What techniques are there to test whether a series converges? Which ones help with this particular series?
 
  • #3
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.
 
  • #4
It is not a homework and I am not a student (and I never was a math student) but I can't prove it. So I will show what I can do :

[itex]\frac{(n+1)^{-2p}}{n^{-2p}} = (\frac{n}{n+1})^{2p} \rightarrow 1[/itex] if [itex]n\rightarrow +\infty[/itex]
We can't conclude.

[itex]\frac{(n+1)^{-2p}}{n^{-2p}} = (\frac{n+1}{n})^{-2p} = (1+\frac{1}{n})^{-2p} = 1-2p\frac{1}{n}+o(\frac{1}{n})[/itex]
[itex]2p>1[/itex] the sum converges.

Is this true?
 
  • #5
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.

p is a positive integer.
 
  • #6
ghc said:
p is a positive integer.

then Yeah the series is convergent for every p in N and your proof seems to be correct to me.
 

FAQ: Convergence of the Series (1/n)^p: A Proof for Positive Integer p

What is the formula for calculating the sum (1/n)^p?

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.

How do you determine if the sum (1/n)^p converges or diverges?

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.

What is the significance of the value of p in the sum (1/n)^p?

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.

Can the sum (1/n)^p converge for all values of p?

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.

Can the sum (1/n)^p converge if p is a negative number?

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.

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