Infinite series of this type converges?

[mathman]

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##?

 

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