'The sum of a series' question

kris1

New member
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
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
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
Hello Kris! And a very warm welcome to the forum...

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
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
Interesting enough, found the same question here before I saw this one : The sum of a series

At least got checked that I answered correct.

kris1

New member
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
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
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
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
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
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 ?