Simple Convergence / Divergence Calc 2

[Alex G]

Homework Statement

I have stared at this too long and do not know which test to approach it with, even writing it out. The problem is
State the Convergence or Divergence of the given series:

Summation n=1 to Infinity of 1 / sqrt (n^3 + 2n)

Homework Equations

I narrowed it down to possibly a comparison test, I can't really see an integral test coming in.
I know the answer in the back of the book says convergence

 

[╔(σ_σ)╝]

Comparison test works well here since sqrt (n^3 + 2n) > n^3/2. The later being a p series with p >1.

Integral test would be hard to use directly without any approximations which can be done with the comparison test.

 

[Alex G]

I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(

 

[Dick]

I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(

If you want to prove convergence, isn't 1/sqrt(n^3+2n)<1/n^(3/2) the condition you want?

 

[╔(σ_σ)╝]

I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(

That's precisely what you want... ;)

 

[Alex G]

/sigh I'm sorry, this is my 6th chapter review section ... I think I'm going to quit while ahead haha, thank you!