Infinite series of this type converges?

[mathman]

TL;DR

Series with $n^{-a}$ convergence is well known for constant a. How about variable $a_n$?

$\sum_{n=1}^\infty n^{-a}$ converge s for $a\gt 1$ - otherwise diverges. Is there any theory for $a_n$? For example $a_n\gt 1$ and $\lim_{n\to \infty} a_n =1$. How about non-convergent with $\liminf a_n=1$?

 

[Mark44]

TL;DR Summary: Series with $n^{-a}$ convergence is well known for constant a. How about variable $a_n$?

$\sum_{n=1}^\infty n^{-a}$ converge s for $a\gt 1$ - otherwise diverges. Is there any theory for $a_n$? For example $a_n\gt 1$ and $\lim_{n\to \infty} a_n =1$. How about non-convergent with $\liminf a_n=1$?

What sort of series are you asking about? Surely it's not $\sum_{n=1}^\infty a_n$, but it's not clear to me what you actually are asking about.

 

[Infrared]

If the limit of the $a_n$ is smaller or larger than 1, then convergence/divergence is clear from comparison. If the limit is 1, then either can happen:

If $a_n=1+1/n$ then $n^{-a_n}=n^{-1-1/n}=n^{-1} n^{-1/n}.$ Since $n^{-1/n}\to 1$ as $n\to\infty$ in this case the series $\sum n^{-a_n}$ diverges by comparison to the Harmonic series.

On the other hand if $a_n=1+2\log\log(n)/\log(n)$ then $n^{-a_n}=n^{-1-2\log\log n/\log n}=n^{-1} n^{-2\log\log n/\log n}=n^{-1} \left(e^{\log n}\right)^{-2\log\log n/\log n}=n^{-1}\log(n)^{-2}.$ Note that $\sum \frac{1}{n\log^2(n)}$ converges (integral test).

 

[Infrared]

What sort of series are you asking about? Surely it's not $\sum_{n=1}^\infty a_n$, but it's not clear to me what you actually are asking about.

I'm pretty sure the OP is replacing the constant $a$ in the sum $\sum n^{-a}$ with a sequence $a_n.$

 

[Mark44]

I'm pretty sure the OP is replacing the constant $a$ in the sum $\sum n^{-a}$ with a sequence $a_n.$

That's what I wanted clarity on. I thought that the OP might mean $\sum n^{a_n}$ but wasn't sure.

 

[pasmith]

I'm pretty sure the OP is replacing the constant $a$ in the sum $\sum n^{-a}$ with a sequence $a_n.$

You can write <br /> \sum n^{-a_n} = \sum e^{-a_n \ln n} so you are in effect analysing series of the form \sum e^{-b_n}. By the ratio test, convergence is then determined by the value of <br /> L = \lim_{n \to \infty} \exp\left(b_n - b_{n+1}\right).