Question

Hey does anyone know series ratio test?? Series enclosed (;

Answer 1

\[\sum_{1}^{\infty} 2^k k!/k^k\]

Answer 2

what have you got so far?

Answer 3

lim as k approaches infnity....

Answer 4

\[2^(k+1) x (k+1)! / (k+1)^(k+1) \times k^k/ 2^k\]

Answer 5

can you tell me this though......what does (k+1)^(k+1) break up into??

Answer 6

that's the rub for me to... let me see if I can think of a way to simplify it

Answer 7

trick one...dos (k+1)! break into K! x 1!?

Answer 8

no, it breaks up into (k+1) x k!
the factorial stuff eliminates nicely, but I'm having a hard time finding a rigorous proof for the limit of k^k/[(k+1)^(k+1)] as k -> infty

Answer 9

ratio test wants us to compute the limit k -> infty of\[{2^{k+1}(k+1)!\over(k+1)^{k+1}}\cdot{k^k\over2^kk!}=2(k+1)\cdot{k^k\over(k+1)^{k+1}}\]

Answer 10

just by eyeballing it it's pretty obvious that \[\lim_{k\to\infty}{k^k\over(k+1)^{k+1}}=0\]since the denominator grows much faster than the top, but I'd like to prove it more rigorously

Answer 11

are you with me up until this point at least?

Answer 12

oh I can see how to finish this off, whenever you'r ready

Answer 13

ready

Answer 14

think about expanding the denominator using the binomial theorem, it will be something like\[(k+1)^{k+1}=k^{k+1}+(k+1)k^k+\cdots\]

Answer 15

so we have\[{2^{k+1}(k+1)!\over(k+1)^{k+1}}\cdot{k^k\over2^kk!}=2(k+1)\cdot{k^k\over k^{k+1}+(k+1)k^k+\frac12(k+1)(k)k^{k-1}+\cdots}\\=2(k+1){k^k\over k^{k+1}+k^{k+1}+k^k+\frac12k^{k+1}+\frac12k^k+\cdots}\]after the binomial expansion, we can ignore all terms but the first few, as they will only serve to make the ratio smaller, so we can just consider the limit\[\lim_{k\to\infty}2(k+1){k^k\over\frac52k^{k+1}+\frac32k^k}=\lim_{k\to\infty}{2k^{k+1}+2k^k\over\frac52k^{k+1}+\frac32k^k}\]

Answer 16

hopefully you can take the limit from here to see that it converges

Answer 17

if you have any questions or get stuck on that limit, let me know