Question

ratio or root test for convergence question

Answer 1

\[\sum_{n=1}^{\infty} (-1)^n \frac{ n^3 }{ 3^n }\]

Answer 2

this is a job for ratio, since there will be wholesale cancellation when you do it

Answer 3

ok let me give it a try and ill ask if i need more help

Answer 4

k

Answer 5

\[\frac{ (-1)^{n+1}\frac{ (n+1)^3 }{ 3^{n+1} } }{ (-1)^n \frac{ n^3 }{ 3^n } }\] having some trouble simplifying this down, i want to just multiply by the recip, but the (-1)* is confusing me on that

Answer 6

whoa whoa hold the phone!

Answer 7

skip that step, and also skip the alternating business
check if absolute values converge forget the \((-1)^n\) part

Answer 8

dividing by a fraction is the same as multiplying by the reciprocal

Answer 9

don't ever write a compound fraction like that if you can help it
\[\frac{(n+1)^3}{3^{n+1}}\times \frac{3^n}{n^3}\]should be your first step

Answer 10

ok

Answer 11

now you see the cancellation clearly right?

Answer 12

if not, let me know
also remember that ratio and root test are for positive terms, i.e. absolute values
not for alternating series,
you are checking if this is absolutely convergent

Answer 13

ok

Answer 14

@satellite73 : Need your assistance to see if you can solve the problem at http://openstudy.com/study#/updates/53550d6ae4b099894ae1dc81. Thanks.