Question

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Posting series now.....

Answer 1

\[\sum_{n=2}^{\infty} (-3n / n+ 1)^{4n}\]

Answer 2

when i do these problems I end up with a whole sheet of paper. When my friend does it, he writes down one equation and says O this is divegent, or o this is convergent ect.... What am I doing wrong and how do I do this problem?

Answer 3

well, the "-" in it might make it troublesome, but thats not always the case; if the function is decreasing in value each term, then it has a better possibility of converging

Answer 4

use the divergence test

Answer 5

if we integrate it as a continuous function, it doesnt converge. I see that much

Answer 6

the innards looks to go to 3 as n got to infinity; so 3^(inf) = inf to me

Answer 7

so it diverges on that account. even tho with the negative itll tend to oscilate back and for

Answer 8

http://mathworld.wolfram.com/DivergenceTests.html

Answer 9

i get \[\infty / \infty \in the limit\]

Answer 10

sorry ment in the limit

Answer 11

\[\lim_{n\to\infty}\left(\frac{-3n}{ n+ 1}\right)^{4n}\neq 0\]

therefore the series diverges

Answer 12

what is the limit then. Just wondering.

Answer 13

you could also use the root test

Answer 14

the limit is infinity

Answer 15

complex infinity i believe

Answer 16

you can't do that

Answer 17

ack!! ... i see it

Answer 18

okay then since the limit doesnt equal zero it diverges. How would I have known to use the test among so many other test.

Answer 19

\[\lim_{n\to\infty}\left(\left[\frac{-3n}{ n+ 1}\right]^4\right)^{n}\]

I like this more

Answer 20

i dont think its about which test to use so much as you have enough tools at your disposal, use them :)

Answer 21

for large n

\[\left[\frac{-3n}{ n+ 1}\right]^4\approx 81\]


\[81^n\] gets big ;)

Answer 22

thanks guys. Ill going to try one more of these then take a breakl lol. Calc make my brain hurt lol.

Answer 23

I am**