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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?

- Thread starter kris1
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- Thread starter
- #1

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?

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- Mar 5, 2012

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Hi kris1! Welcome to MHB!

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?

Suppose we pick an $a_n$ so that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converges, but only

Say, \(\displaystyle a_n^2 = \frac 1 {n^{7/6}}\)...

- Jan 17, 2013

- 1,667

Your choice of $a_n$ doesn't contradict the convergence of \(\displaystyle \sum_{n\geq 2}\frac{a_n}{n^{2/3}}\)Hi kris1! Welcome to MHB!

Suppose we pick an $a_n$ so that \(\displaystyle \sum_{n=1}^{\infty} a^2_{n}\) converges, but onlyjust.

Say, \(\displaystyle a_n^2 = \frac 1 {n^{7/6}}\)...

- Sep 16, 2013

- 337

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, \(\displaystyle a_n/ n^z \le a_n^2\)...

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- Feb 7, 2012

- 2,702

Hint: Use the Cauchy–Schwarz inequality.

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?

- Mar 22, 2013

- 573

At least got checked that I answered correct.

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- #7

- Mar 22, 2013

- 573

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

Can you prove this now?

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- #10

- Mar 5, 2012

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It looks like you really need the Cauchy-Schwarz inequality... in combination with the

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- #11

- Feb 7, 2012

- 2,702

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.

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