Question

\[\sum_{k=1}^{\infty} \frac{2^{k}k!}{(k+2)!}\]
\[\frac{2^{k}}{(k+2)}\]

am I on the right track?

Answer 1

ratio test?

Answer 2

final answer 4/3

Answer 3

yes

Answer 4

\[\frac{2^{(k+1)}}{2^k} \frac{k+2}{k+3}\]

Answer 5

i got
\[\frac{2(k+1)}{k+3}\] but i could be wrong

Answer 6

you want to do it slowly?

Answer 7

I got to that part too then i divided by k

Answer 8

both numerator and denominator

Answer 9

? why

Answer 10

to let k-> \[\infty\]

Answer 11

i suppose you could but lets think
\[\frac{2(k+1)}{k+3}=\frac{2k+6}{k+3}\] numerator is a polynomial of degree 1 as is the denominator
since the degrees are the same, the limit as \(k\to \infty\) is the ratio of the leading coefficients, namely 2
i.e.
\[\lim_{k\to \infty}\frac{2k+6}{k+3}=\frac{2}{1}=2\]
that part you should do in your head from an old pre - calc class

Answer 12

2(k+1)=(2k +2)

Answer 13

lol so it is

Answer 14

of course it make absolutely no difference

Answer 15

not 2k +6 lolz....so 2/3

Answer 16

noooo

Answer 17

convergent?

Answer 18

has nothing at all to do with the constant, just ratio of the leading coefficients

Answer 19

no divergent, but i have confused you i sense

Answer 20

so the final answer is 2/3 which is less than 1 so it's convergent

Answer 21

do you remember a pre calc class of some kind where you looked for horizontal asymptotes?

Answer 22

if the degree of the numerator is the same as the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients

Answer 23

you have
\[\lim_{k\to \infty}\frac{2k+2}{k+3}=\frac{2}{1}=2\] so diverges

Answer 24

forget the constants, they have nothing to do with it

Answer 25

yes. but here is my point though

\[\lim_{k\to \infty}\frac{2k+6}{k+3}=\frac{2}{1}=2\]
\[\lim_{k\to \infty}\frac{2k+1}{k+3}=\frac{2}{3}=\frac{2}{3}\]

Answer 26

\lim_{k\to \infty}\frac{2(k+1)}{k+3}=\frac{2}{3}=\frac{2}{3}

Answer 27

oh no it has nothing to do with the constant
\[\lim_{k\to \infty}\frac{2k+555555}{k+1}=2\]

Answer 28

\[\lim_{k\to \infty}\frac{2(k+1)}{k+3}=\frac{2}{3}=\frac{2}{3}\]

Answer 29

the constant is not important in this calculation, it is only the ratio of the leading coefficients

Answer 30

ok....I must accept that fact then

Answer 31

this line is incorrect
\[\lim_{k\to \infty}\frac{2(k+1)}{k+3}=\frac{2}{3}=\frac{2}{3}\] but you do not have to just accept it, you can make sense out of it as follows

Answer 32

imagine \(k\) is 1,000,000
then numerator is 2,000,002 and the denominator is 1,000,003

Answer 33

ok

Answer 34

when you divide, you do not get 2 exactly but it is very very close to 2 right? the constant is unimportant in this calculation because k goes to infinity, i.e. gets very large and the constant is irrelevant

Answer 35

yes

Answer 36

good! so to finish this problem we see by the ratio test that the limit is 2 and therefore it diverges

Answer 37

makes sense

Answer 38

yay