Infinite series of this type converges?

In summary, the convergence of series with terms ##n^{-a}## is well known for a constant value of ##a##, but what about when the value of ##a## varies as a sequence ##a_n##? By replacing ##a## with a sequence, the series becomes ##\sum n^{-a_n}## and can be analyzed using the ratio test. The convergence of this series is determined by the limit of ##\exp(b_n - b_{n+1})##, where ##b_n## is the corresponding value of ##a_n##.
  • #1
mathman
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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##?
 
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  • #2
mathman said:
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.
 
  • #3
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).
 
  • #4
Mark44 said:
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.##
 
  • #5
Infrared said:
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.
 
  • #6
Infrared said:
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]
 

What is an infinite series?

An infinite series is a sum of an infinite number of terms, where each term is related to the previous one by a specific pattern or rule.

How do you determine if an infinite series converges?

To determine if an infinite series converges, you can use various tests such as the comparison test, ratio test, or integral test. These tests analyze the behavior of the terms in the series to see if they approach a finite limit or not.

What does it mean for an infinite series to converge?

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms increases. This means that the series has a finite and well-defined value.

What happens if an infinite series does not converge?

If an infinite series does not converge, it is said to diverge. This means that the sum of its terms does not approach a finite limit and the series has no well-defined value.

Can an infinite series converge to a negative value?

Yes, an infinite series can converge to a negative value. The sign of the sum of the terms in a series is not indicative of whether it converges or not. It is the behavior of the terms that determines convergence or divergence.

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