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'The sum of a series' question

kris1

New member
Dec 4, 2013
4
Assume that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converge, and assume that \(\displaystyle a_{n}\) is non-negative for all \(\displaystyle \textit{n} \in N.\)
Determine whether the following statement is true (and prove it) or false (and give counterexample).
\(\displaystyle \sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty\)

Does anyone know how to do this question?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Assume that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converge, and assume that \(\displaystyle a_{n}\) is non-negative for all \(\displaystyle \textit{n} \in N.\)
Determine whether the following statement is true (and prove it) or false (and give counterexample).
\(\displaystyle \sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty\)

Does anyone know how to do this question?
Hi kris1! Welcome to MHB! :)

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hi kris1! Welcome to MHB! :)

Suppose we pick an $a_n$ so that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converges, but only just.
Say, \(\displaystyle a_n^2 = \frac 1 {n^{7/6}}\)...
Your choice of $a_n$ doesn't contradict the convergence of \(\displaystyle \sum_{n\geq 2}\frac{a_n}{n^{2/3}}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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 \(\displaystyle a_n^2\) converges, then any sum of fractions of powers less than \(\displaystyle a_n^2 \) should also converge, since all you need to do is show that, say, \(\displaystyle a_n/ n^z \le a_n^2\)...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Assume that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converge, and assume that \(\displaystyle a_{n}\) is non-negative for all \(\displaystyle \textit{n} \in N.\)
Determine whether the following statement is true (and prove it) or false (and give counterexample).
\(\displaystyle \sum_{n=2}^{\infty} \frac{a_{n}}{n^{2/3}}<\infty\)

Does anyone know how to do this question?
Hint: Use the Cauchy–Schwarz inequality.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Interesting enough, found the same question here before I saw this one : The sum of a series :p

At least got checked that I answered correct. :D
 

kris1

New member
Dec 4, 2013
4
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!
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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?
 

kris1

New member
Dec 4, 2013
4
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 :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 \(\displaystyle \frac{a_n}{n^{2/3}} \leqslant \frac12\Bigl(a_n^2 + \frac1{n^{4/3}}\Bigr).\) Then \(\displaystyle \sum \frac{a_n}{n^{2/3}} \leqslant \frac12\Bigl(\sum a_n^2 +
\sum \frac1{n^{4/3}}\Bigr)\), and it follows from the comparison test that \(\displaystyle \sum \frac{a_n}{n^{2/3}}\) converges.
 

kris1

New member
Dec 4, 2013
4
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 \(\displaystyle a_{n}\)=2, and n=8 ?