Does Convergence of a Series Imply Convergence of Its Terms Over n?

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    Convergence Series
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Homework Help Overview

The discussion revolves around the convergence of series, specifically exploring whether the convergence of the series \(\sum a_n\) implies the convergence of the series \(\sum \frac{a_n}{n}\) when \(a_n \geq 0\). Participants are analyzing the implications of convergence and the relationships between different series formed from the terms \(a_n\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the comparison test but expresses difficulty in establishing a suitable \(N_0\) for the inequality involving \(\frac{\sqrt{a_n}}{n}\) and \(a_n\). Other participants suggest using partial sums and reference known inequalities, such as Cauchy-Schwarz, to further their reasoning.

Discussion Status

Participants are actively engaging with the problem, sharing insights and corrections regarding notation. There is a collaborative effort to clarify the approach and explore the implications of the convergence of different series. No explicit consensus has been reached, but productive lines of inquiry are being pursued.

Contextual Notes

There is a mention of the inability to edit the original post, which may affect the clarity of the problem statement. The discussion also highlights potential confusion regarding variable notation.

ehrenfest
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[SOLVED] convergence of a series

Homework Statement


Prove that the convergence of [itex]\sum a_n[/itex] implies the convergence of [itex]\sum \frac{a_n}{n}[/itex] if [itex]a_n \geq 0[/itex].

Homework Equations


The Attempt at a Solution


I want to use the comparison test. So, I want to find [itex]N_0[/itex] so that [itex]n \geq N_0[/itex] implies [tex]\frac{\sqrt{a_n}}{n} \leq a_n[/tex] which is clearly not in general possible. So I am stuck. Maybe I need to replace a_n by a subsequence of a_n or something?
 
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consider any partial sum A_m = Sum(a_k/k, k = 1,..., m).

|A_m| = |a_1/1 + ... + a_k/k| = |(1/1) * a_1 + ... + (1/k)*a_k| <= ...
 
Sorry. I am trying to prove
[tex] \sum \frac{\sqrt{a_n}}{n}[/tex]

converges not

[tex] \sum \frac{a_n}{n}[/tex]

For some reason, I cannot edit the opening post.
 
ok even easier, just means you won't have to show something else I had in mine,

Let A_m = sum(sqrt(a_k)/k, k = 1,..., m) be any partial sum, then

|A_m| = |sqrt(a_1)/1 + ... + sqrt(a_k)/k| = |(1/1)*sqrt(a_1) + ... + (1/k)sqrt(a_k)| <= ...

use some famous inequality for the next step(not the triangle inequality), what's the other one I'm sure you know!
 
Cauchy-Schwarz. Yay!

BTW you're k's and m's are mixed up.
 
ehrenfest said:
Cauchy-Schwarz. Yay!

BTW you're k's and m's are mixed up.

woops yea, good to see you got it! I'll look at your other question now
 

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