Question

does the infinite series (n+1)(n+6)/n! converge or diverge? which test should I use to find out?

Answer 1

ratio test is convenient to use with factorial terms

Answer 2

it doesn't cancel out though: I end up with (n+2)(n+7)*(n+1)/(n+1)!(n+1)(n+6)

Answer 3

(n+1)! = n!(n+1)

Answer 4

but I still have n! left in the denominator

Answer 5

ok so I have just a polynomial over a polynomial with the same degree. the limit would be 1. that doesn't help me in the ratio test

Answer 6

no...

Answer 7

\[\frac{(n+2)(n+7) }{n!(n+1) } \frac{ n! }{ (n+6)(n+1)}\]

Answer 8

I got up till there, I just can't simplify it in any meaningful way

Answer 9

the limit is zero

Answer 10

how come?

Answer 11

~n^2 / n^3

Answer 12

aka 1/n

Answer 13

!!!! yes!! I think I forgot to multiply the second (n+1) and that's why I had n^2/n^2

Answer 14

so it converges because the limit is less than 1?

Answer 15

yes

Answer 16

thank you!

I have one more question: I have (1-p/k)^k and I have to find for what p does it converge

Answer 17

I would be taking the limit of 1/e^p right?

Answer 18

( (1-p)/k )^k ?

Answer 19

(1-(p/k))^k