infinite series 2

[mario99]

Determine convergence or divergence of the series.

\(\displaystyle \sum_{k=1}^{\infty} \frac{4}{\sqrt[3]{k}}\)

Any help would be appreciated!

 

[Deleted member 4993]

Determine convergence or divergence of the series.

\(\displaystyle \sum_{k=1}^{\infty} \frac{4}{\sqrt[3]{k}}\)

Any help would be appreciated!

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Dr.Peterson]

Determine convergence or divergence of the series.

\(\displaystyle \sum_{k=1}^{\infty} \frac{4}{\sqrt[3]{k}}\)

Any help would be appreciated!

I see that you have a history of never showing (or doing?) any work, but only "borrowing" the work of others. You need to change that before you'll get any help here.

 

[mario99]

Thank you very much Subhotosh Khan and Dr.Peterson for helping me.

Don't worry about the history. At first, it seems like I am taking an exam and I am cheating but later you will discover that it is just a matter of an style to throw a bunch of problems at once and then have a discussion.

I have never borrowed the work of others but I have thanked others for getting me started.

This infinite series converges because it is different than the famous \(\displaystyle \sum_{k=1}^{\infty}\frac{4}{k}\). This is my first guess.

 

[Dr.Peterson]

This infinite series converges because it is different than the famous \(\displaystyle \sum_{k=1}^{\infty}\frac{4}{k}\). This is my first guess.

What tests have you tried? Those work better than guesses!

Have you learned about p-series?

This may help:

Calculus II - Integral Test

In this section we will discuss using the Integral Test to determine if an infinite series converges or diverges. The Integral Test can be used on a infinite series provided the terms of the series are positive and decreasing. A proof of the Integral Test is also given.

tutorial.math.lamar.edu

 

[pka]

Determine convergence or divergence of the series. \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{\sqrt[3]{k}}\)!

[imath]\displaystyle\sum\limits_{k = 1}^\infty {\frac{4}{{\sqrt[4]{k}}}} = 4\sum\limits_{k = 1}^\infty {\frac{1}{{\sqrt[4]{k}}}} [/imath] Now if [imath]n\in\mathbb{Z}^+[/imath] then [imath]\sqrt[n]{n} \leqslant n \Rightarrow \dfrac{1}{n} \leqslant \dfrac{1}{{\sqrt[n]{n}}}[/imath][imath][/imath]
Now apply the basic comparison test. See the attached.

 

[Steven G]

Thank you very much Subhotosh Khan and Dr.Peterson for helping me.

Don't worry about the history. At first, it seems like I am taking an exam and I am cheating but later you will discover that it is just a matter of an style to throw a bunch of problems at once and then have a discussion.

I have never borrowed the work of others but I have thanked others for getting me started.

This infinite series converges because it is different than the famous \(\displaystyle \sum_{k=1}^{\infty}\frac{4}{k}\). This is my first guess.

No one accused you of submitting test problems and then waiting to receive the solutions to your problems.
On this forum we never answer questions for students. We rather discuss the problem with the student but FIRST the student needs to show some work.

 

[mario99]

Thank you very much pka and Jomo for helping me.

What tests have you tried? Those work better than guesses!

Have you learned about p-series?

This may help:

Calculus II - Integral Test

In this section we will discuss using the Integral Test to determine if an infinite series converges or diverges. The Integral Test can be used on a infinite series provided the terms of the series are positive and decreasing. A proof of the Integral Test is also given.

tutorial.math.lamar.edu

Tons of tests. The link that about the integral test that you have sent is easy to use but I can't figure out when the function will be decreasing. This is a crucial part to use the test.

The test tells the series will diverge since \(\displaystyle \int_1^{\infty} \frac{4}{\sqrt[3]{x}} \ dx = \infty\).

The p-series was the first test to learn.

[imath]\displaystyle\sum\limits_{k = 1}^\infty {\frac{4}{{\sqrt[4]{k}}}} = 4\sum\limits_{k = 1}^\infty {\frac{1}{{\sqrt[4]{k}}}} [/imath] Now if [imath]n\in\mathbb{Z}^+[/imath] then [imath]\sqrt[n]{n} \leqslant n \Rightarrow \dfrac{1}{n} \leqslant \dfrac{1}{{\sqrt[n]{n}}}[/imath][imath][/imath]
Now apply the basic comparison test. See the attached.

This is true. \(\displaystyle \frac{1}{n} \leq \frac{1}{\sqrt[n]{n}}\)
But how will I know if this is always true? \(\displaystyle \frac{1}{n} \leq \frac{1}{\sqrt[3]{n}}\)

If this is the case this will allow us to see that the series diverges since \(\displaystyle \sum_{k=1}^{\infty}\frac{4}{k} \leq \sum_{k=1}^{\infty}\frac{4}{\sqrt[3]{k}}\).

The problem is that I will not be able to come with this idea \(\displaystyle \sqrt[k]{k} \leq k\).

No one accused you of submitting test problems and then waiting to receive the solutions to your problems.
On this forum we never answer questions for students. We rather discuss the problem with the student but FIRST the student needs to show some work.

Maybe you should take a tour on my last \(\displaystyle 10\) posts to see that.

 

[Dr.Peterson]

Tons of tests. The link that about the integral test that you have sent is easy to use but I can't figure out when the function will be decreasing. This is a crucial part to use the test.

The test tells the series will diverge since \(\displaystyle \int_1^{\infty} \frac{4}{\sqrt[3]{x}} \ dx = \infty\).

The p-series was the first test to learn.

That page also shows the p test, near the bottom, which directly tells you that since p = 1/3 < 1, the series diverges.

One way to tell whether a function is increasing or decreasing is to take its derivative. Did you try that?

 

[mario99]

\(\displaystyle \frac{d}{dx}(\frac{4}{\sqrt[3]{x}}) = -\frac{4}{3x^{4/3}}\)

Does the negative imply that the function is decreasing?

 

[Deleted member 4993]

\(\displaystyle \frac{d}{dx}(\frac{4}{\sqrt[3]{x}}) = -\frac{4}{3x^{4/3}}\)

Does the negative imply that the function is decreasing?

What does your textbook say?

Also plot your function using WolframAlpha.com and you can visualize the increasing-decreasing zones.

 

[BigBeachBanana]

\(\displaystyle \frac{d}{dx}(\frac{4}{\sqrt[3]{x}}) = -\frac{4}{3x^{4/3}}\)

Does the negative imply that the function is decreasing?

Is it always negative? Try x=1 and x=-1.

 

[Dr.Peterson]

\(\displaystyle \frac{d}{dx}(\frac{4}{\sqrt[3]{x}}) = -\frac{4}{3x^{4/3}}\)

Does the negative imply that the function is decreasing?

Not the negative sign, but the fact that the value is negative (because, in context, you are only interested in values of x greater than 1).