# Infinite series of this type converges?

• A
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##?

## Answers and Replies

Mentor
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.

Gold Member
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).

Gold Member
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.##

Mentor
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.

Homework Helper
2022 Award
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 $$\sum n^{-a_n} = \sum e^{-a_n \ln n}$$ 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 $$L = \lim_{n \to \infty} \exp\left(b_n - b_{n+1}\right).$$