'The sum of a series' question

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Discussion Overview

The discussion revolves around the convergence of the series $$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}$$ given that $$\sum_{n=1}^{\infty} a^2_{n}$$ converges and that $$a_{n}$$ is non-negative. Participants explore whether the statement is true or false, and they discuss various approaches to prove or disprove it, including the use of the Cauchy-Schwarz inequality and convergence tests.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that if $$\sum_{n=1}^{\infty} a^2_{n}$$ converges, then the series $$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}$$ should also converge, proposing that the terms of the latter series are dominated by the former.
  • Others argue that specific choices of $$a_n$$, such as $$a_n^2 = \frac{1}{n^{7/6}}$$, could lead to different conclusions about the convergence of $$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}$$.
  • A participant mentions the Cauchy-Schwarz inequality as a potential method to prove the convergence of the series, suggesting that it can be applied to relate the two series.
  • Another participant expresses uncertainty about using the Cauchy-Schwarz inequality, indicating that they have only learned other convergence tests and feel unprepared to apply this method.
  • One participant provides a detailed application of the Cauchy-Schwarz inequality to show how it can lead to a conclusion about the convergence of the series.
  • There is a mention of the Direct Comparison Test in conjunction with the Cauchy-Schwarz inequality as a way to establish convergence.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the convergence of the series $$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}$$. While some propose methods to prove convergence, others raise counterexamples and express uncertainty about the applicability of certain techniques.

Contextual Notes

Some participants note limitations in their understanding of the Cauchy-Schwarz inequality and other convergence tests, which may affect their ability to engage fully with the problem.

kris11
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Assume that $$\sum_{n=1}^{\infty} a^2_{n}$$ converge, and assume that $$a_{n}$$ is non-negative for all $$\textit{n} \in N.$$
Determine whether the following statement is true (and prove it) or false (and give counterexample).
$$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty$$

Does anyone know how to do this question?
 
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kris1 said:
Assume that $$\sum_{n=1}^{\infty} a^2_{n}$$ converge, and assume that $$a_{n}$$ is non-negative for all $$\textit{n} \in N.$$
Determine whether the following statement is true (and prove it) or false (and give counterexample).
$$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty$$

Does anyone know how to do this question?

Hi kris1! Welcome to MHB! :)

Suppose we pick an $a_n$ so that $$\sum_{n=1}^{\infty} a^2_{n}$$ converges, but only just.
Say, $$a_n^2 = \frac 1 {n^{7/6}}$$...
 
I like Serena said:
Hi kris1! Welcome to MHB! :)

Suppose we pick an $a_n$ so that $$\sum_{n=1}^{\infty} a^2_{n}$$ converges, but only just.
Say, $$a_n^2 = \frac 1 {n^{7/6}}$$...

Your choice of $a_n$ doesn't contradict the convergence of $$\sum_{n\geq 2}\frac{a_n}{n^{2/3}}$$
 
Hello Kris! And a very warm welcome to the forum... (Hug)I'm not much of a series boffin, but it seems to me that if an infinite sum of squares $$a_n^2$$ converges, then any sum of fractions of powers less than $$a_n^2 $$ should also converge, since all you need to do is show that, say, $$a_n/ n^z \le a_n^2$$...
 
kris1 said:
Assume that $$\sum_{n=1}^{\infty} a^2_{n}$$ converge, and assume that $$a_{n}$$ is non-negative for all $$\textit{n} \in N.$$
Determine whether the following statement is true (and prove it) or false (and give counterexample).
$$\sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty$$

Does anyone know how to do this question?
Hint: Use the Cauchy–Schwarz inequality.
 
Interesting enough, found the same question here before I saw this one : http://www.mymathforum.com/viewtopic.php?f=22&t=44756 :p

At least got checked that I answered correct. :D
 
Hi Everyone :) Thank you for all your replies :) I still haven't done this question :( I don't know what Cauchy–Schwarz inequality is, but I looked for this inequality on the internet and tried to apply it to my question but it didn't work. I also tried convergence tests but I gave up. I can do that kind of questions but when I have real numbers and I have to find limits, but for this question I need to prove given statement, which is my greatest weakness :( Please help me!
 
Applying CS-inequality,

$$\left ( \sum_{n\geq2} \frac{a_n}{n^{2/3}}\right )^2 \leq \left (\sum_{n \geq 2} a_n^2 \right ) \left (\sum_{n\geq 2} \frac{1}{n^{4/3}} \right ) $$

Can you prove this now?
 
Yeah, I think I can prove it, however I cannot use this method as I never learn it before. So far for this kind of questions I learned tests for convergence, eg ratio test, root test, integral test... etc :) So I think my teacher wants me to solve this question using these methods. But thank you very much for your replies :)
 
  • #10
kris1 said:
Yeah, I think I can prove it, however I cannot use this method as I never learn it before. So far for this kind of questions I learned tests for convergence, eg ratio test, root test, integral test... etc :) So I think my teacher wants me to solve this question using these methods. But thank you very much for your replies :)

It looks like you really need the Cauchy-Schwarz inequality... in combination with the Direct Comparison Test.
 
  • #11
kris1 said:
Yeah, I think I can prove it, however I cannot use this method as I never learn it before. So far for this kind of questions I learned tests for convergence, eg ratio test, root test, integral test... etc :) So I think my teacher wants me to solve this question using these methods. But thank you very much for your replies :)
If $x$ and $y$ are positive numbers then $x^2 + y^2 - 2xy =(x-y)^2 \geqslant0$, from which $xy \leqslant \frac12(x^2+y^2).$ Apply that with $x=a_n$ and $y = n^{-2/3}$ to see that $$\frac{a_n}{n^{2/3}} \leqslant \frac12\Bigl(a_n^2 + \frac1{n^{4/3}}\Bigr).$$ Then $$\sum \frac{a_n}{n^{2/3}} \leqslant \frac12\Bigl(\sum a_n^2 +
\sum \frac1{n^{4/3}}\Bigr)$$, and it follows from the comparison test that $$\sum \frac{a_n}{n^{2/3}}$$ converges.
 
  • #12
Ohh thank you! So for the first part you used Cauchy-Schwarz inequality, right? And can I prove it by using real numbers, eg for $$a_{n}$$=2, and n=8 ?
 

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