series

[Brain0991]

Test the series below for convergence or divergence.

n=0 to infinity [n/(3n+2)]

I know there is some rule for limits where if the powers are the same on the top and the bottom you take the coefficients, but that does not really prove anything here I guess. I would use a comparison test but I am not sure what to compare it to.

Is the Integral test my only option?

 

[Brain0991]

Would it prove anything if i took the limit as n-->infinity of a[sub:2h9pdd6h]n[/sub:2h9pdd6h]?

would it not just be infinity/infinity and since it is not 0 the series diverges?

 

[Deleted member 4993]

Would it prove anything if i took the limit as n-->infinity of a[sub:14n36wu8]n[/sub:14n36wu8]?

would it not just be infinity/infinity and since it is not 0 the series diverges?

You must be taking very long naps during the class!!

What condition does a[sub:14n36wu8]n[/sub:14n36wu8] must satisfy for convergence?

 

[Brain0991]

is it that obvious!? jk had a brain fart. way too many different rules of series and what-not running through my head

The limit as n--> infinity of a[sub:1uz7lbxu]n[/sub:1uz7lbxu] must be equal to 0 for the series to be convergent

and since you get it to be the limit of [n/(3n+2)]and the highest powers are the same on the top and bottom you can take the coefficients which would be 1/3 which is not zero therefore the series is DIVERGENT??

 

[Deleted member 4993]

is it that obvious!? jk had a brain fart. way too many different rules of series and what-not running through my head

The limit as n--> infinity of a[sub:36w59ngu]n[/sub:36w59ngu] must be equal to 0 for the series to be convergent (remember this is a necessary condition - not sufficient)

and since you get it to be the limit of [n/(3n+2)]and the highest powers are the same on the top and bottom you can take the coefficients which would be 1/3 which is not zero therefore the series is DIVERGENT?? Correct