A logarithmic convergence tests - Analysis

Click For Summary

Homework Help Overview

The discussion centers around the convergence of a series involving a non-negative sequence \{a_{n}\}_{n=1}^{\infty} and its relationship with the series of logarithms, specifically \(\Sigma_{n=1}^{\infty}a_{n}\) and \(\Sigma_{n=1}^{\infty}\ln(1+a_{n})\). Participants are exploring the conditions under which these series converge or diverge.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • One participant attempts to apply the direct comparison test but struggles to find a suitable constant γ. Others discuss the transformation of the logarithmic series into a product and question how this relates to the convergence of the original series. There are inquiries about the implications of the convergence of \(\Sigma_{n=1}^{\infty}\ln(1+a_{n})\) on \(\Sigma_{n=1}^{\infty}a_{n}\) and vice versa.

Discussion Status

The discussion is active, with participants raising questions about the relationships between the series and exploring different mathematical properties. Hints have been provided to guide the exploration of logarithmic properties and their implications for convergence, but no consensus or resolution has been reached.

Contextual Notes

Participants are working under the constraints of proving convergence without providing complete solutions, focusing on the logical connections between the series involved.

Wuberdall
Messages
32
Reaction score
0

Homework Statement



Given a non-negative sequence [itex]\{a_{n}\}_{n=1}^{\infty}[/itex]. Proove that the serie [itex]\Sigma_{n=1}^{\infty}a_{n}[/itex] converge if and only if [itex]\Sigma_{n=1}^{\infty}\ln(1+a_{n})[/itex] converges.



Homework Equations





The Attempt at a Solution



My first attempt is the direct comparison test. Which tells that if [itex]\Sigma_{n=1}^{\infty}\ln(1+a_{n})[/itex] converges and if there exist a reel number γ so [itex]a_{n}\leq\gamma\ln(1+a_{n})[/itex] for all n, then the series [itex]\Sigma_{n=1}^{\infty}a_{n}[/itex] also converges.

But i can't find such a γ.
 
Last edited:
Physics news on Phys.org
Hint*: Note that ##\sum_{n=1}^\infty ln(1+a_n)## is a sum of logarithms. What can you do to change logarithms into multiplication?

*##log(a)+log(b)=log(ab)##.
 
yeah, so [itex]\Sigma_{n=1}^{\infty}\ln(1+a_{n}) = \ln\Big[\Pi_{n=1}^{\infty}(1+a_{n})\Big][/itex]. But how am I supposed to conclude from this, that the series [itex]\Sigma_{n=1}^{\infty}a_{n}[/itex] must then also converge ?
 
Well we know that ##\sum_{n=1}^\infty ln(1+a_n)=C##, where C is a constant, since the sum converges. And we know ##\sum_{n=1}^\infty ln(1+a_n)=ln[\prod_{n=1}^\infty(1+a_n)]=C##. So if we take the exponent of both sides we get ##\prod_{n=1}^\infty(1+a_n)=e^C##. If you expand the product what do you notice? Now you have a bound for the sum ##\sum_{n=1}^\infty a_n ##
 
Wuberdall said:
yeah, so [itex]\Sigma_{n=1}^{\infty}\ln(1+a_{n}) = \ln\Big[\Pi_{n=1}^{\infty}(1+a_{n})\Big][/itex]. But how am I supposed to conclude from this, that the series [itex]\Sigma_{n=1}^{\infty}a_{n}[/itex] must then also converge ?

Why is that if every [itex]a_n \geq 0[/itex], we have that [tex]\exp\left(\sum_{n=1}^N \ln(1 + a_n)\right) = \prod_{n=1}^N (1 + a_n) \geq 1 + \sum_{n=1}^N a_n \geq 1[/tex] for every [itex]N \in \mathbb{N}[/itex]? (Hint: consider what happens if you expand the product.)

If [itex]\sum_{n=1}^\infty \ln(1 + a_n)[/itex] converges, what can you say about the convergence of [itex]\sum_{n=1}^\infty a_n[/itex]?

Conversely, if [itex]\sum_{n=1}^\infty a_n[/itex] diverges, what can you conclude about the convergence of [itex]\sum_{n=1}^\infty \ln(1 + a_n)[/itex]?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K