Question

HELP! CALCULUS - CONVERGENCE/DIVERGENCE of SERIES: sum to infinity from n = 1 (2^(n+1))/(n^n)

Answer 1

do i use the ratio test?

Answer 2

Converges

Answer 3

to prove that, first, realize that when an exponent has a +1 it just means you multiply by the base... so (2^(n+1))/(n^n)=2(2^(n))/(n^n)

Answer 4

then use nth root test

Answer 5

i can see why you thought of ratio test, because you saw the n+1 and the n, but you must notice that they have different bases and if you conducted the ratio test it would have been like this: ((2^(n+1))/(n^n))/(2^(n+2))/((n+1)^(n+1)) ... which is very ugly :)

Answer 6

err, inverse that, my bad... (2^(n+2))/((n+1)^(n+1))/((2^(n+1))/(n^n))...

Answer 7

still ugly

Answer 8

cheers @arman1002 @jlvm

Answer 9

hmm? cheers...? to what (if i may ask?)

Answer 10

lol. in australia cheers = thanks

Answer 11

i see, your from australia! thats pretty cool. probably the one place i havent been