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Convergence of a series

  • Thread starter ehrenfest
  • Start date
1,996
1
[SOLVED] convergence of a series

1. 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].


2. Homework Equations



3. 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?
 

Answers and Replies

229
0
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| <= ...
 
1,996
1
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.
 
229
0
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!
 
1,996
1
Cauchy-Schwarz. Yay!

BTW you're k's and m's are mixed up.
 
229
0
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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