Question

Determine whether the series converges or diverges:

Answer 1

\[\sum_{}^{}\frac{ k+1 }{ \sqrt{2k^3+k} }\]

Answer 2

@ganeshie8

Answer 3

I think I need to use limit comparison test?

Answer 4

yes, but which bn ?

Answer 5

This is a picture of how one of my classmates did the question, but I dont understand some of his steps.

Answer 6

\[\frac{ 1 }{ k^3 }\]
?

I say that because I think you usually want to pick the highest power right?

Answer 7

Your classmate divided each term by k.

Answer 8

why?

Answer 9

and
\[\frac{ 2k^3 }{ k } \neq 2k\]

So that part in the denominator is confusing me too

Answer 10

if you divide all terms by k you get
\[ \frac{1+\frac{1}{k}}{\sqrt{2k + \frac{1}{k}} }\]
maybe consider
1/ sqrt(2k) ?
we know 1/k diverges

Answer 11

Sorry if im stuck on the basics here.

is the denominator \[\sqrt{2k+\frac{ 1 }{ k }}\]

because

\[\sqrt{2k^3+k}= \sqrt{2k(k^2+1)}\]?

Im not sure how he divided the stuff in the radical by k :/

Answer 12

use k= sqrt(k^2)

Answer 13

oh oh course

Answer 14

so \[ \frac{\sqrt{2k^3+k}}{k} = \sqrt{ \frac{2k^3+k}{k^2} }\]

Answer 15

I think it's clear it diverges. Just need the details

Answer 16

Yea. Can I use basic divergence test at this point?

Answer 17

Or do I need to use P Series

Answer 18

See if below manipulation looks any simpler :

\[\frac{ k+1 }{ \sqrt{2k^3+k} } \ge\frac{ k+1 }{ \sqrt{2k^3+2k^3} } \ge \dfrac{k}{2k\sqrt{k}} = \dfrac{1}{2\sqrt{k}} \]

Answer 19

That makes sense @ganeshie8

Answer 20

See if below manipulation looks any simpler :

\[\frac{ k+1 }{ \sqrt{2k^3+k} } \ge\frac{ k+1 }{ \sqrt{2k^3+2k^3} } = \dfrac{k+1}{\sqrt{4k^3}} \ge \dfrac{k}{2k\sqrt{k}} = \dfrac{1}{2\sqrt{k}} \]

Answer 21

So really the trick here is just rewriting using algebra and then using p series

Answer 22

It looks like the trick is to do a bunch of these and learn the techniques.
Otherwise, you will be like me and wander about in search of an idea.

Answer 23

We are in the same Boat @phi lol

Answer 24

@ganeshie8 did you replace k with 2k^3 just so it would be easy to pull out of the radical?

Answer 25

Yes, but why are we allowed to replace like that ?

Answer 26

Yes - In college I learned about 20 different tests for convergence but I cant access them in my brain anymore!!

Answer 27

I dont know. I thought you could if the values converged, but this diverges.

Answer 28

Haha there are more than a dozen convergence tests. I remember only the comparison test ..

Answer 29

And the geometric series, ofcourse !

Answer 30

yea

Answer 31

Yea, Im really hoping I dont have to remember them next semester in Calc 3

Answer 32

Let me break it into few steps for you

Answer 33

Would you agree that \(k\le 2k^3\) is true for all \(k\gt 1\) ?

Answer 34

yea for sure

Answer 35

Add \(2k^3\) both sides and get

\(k + \color{red}{2k^3} \le 2k^3+ \color{red}{2k^3} \)

Answer 36

Still with me ?

Answer 37

yes

Answer 38

Take square root both sides and get

\(\sqrt{k + \color{red}{2k^3} }\le \sqrt{2k^3+ \color{red}{2k^3}} \)

Answer 39

Next, take flip both sides. What happens to the inequality when you flip ?

Answer 40

It also flips

Answer 41

@legomyego180 Calc3 is mostly about line integrals and surface integrals. But it helps to remember comparison test and geometric series. Some proofs in Calc3 may use these.

Answer 42

Flipping both sides gives

\(\dfrac{1}{\sqrt{k + \color{red}{2k^3} }}\ge \dfrac{1}{\sqrt{2k^3+ \color{red}{2k^3}} }\)

Answer 43

Multiply k+1 both sides

Answer 44

I see now. Because \[\frac{ k+1 }{ \sqrt{2k^3+3} }\] is greater than \[\frac{ 1 }{ 2k^3+2k^3 }\]
(forgot the square root sign)

And because \[\frac{ 1 }{ \sqrt{2k^3+2k^3} }\] diverges
that means our original equation must also diverge because it is greater.