Question

Show that the following sequences converge n!/n^n

Answer 1

\[\frac{ n! }{ n^n }\]

Answer 2

Would be great if you could use Sterling's formula here.

Answer 3

never heard of it. maybe squeeze therom.

Answer 4

What you wanna bound it above by?

Answer 5

ill upload a pic, of what the lecturer posted. i dont understand the upper bound. 1 sec

Answer 6

here: http://i.imgur.com/nlQwQxS.png

Answer 7

Wait, this isn't series convergence... just limit convergence?

Answer 8

not sure, what u mean

Answer 9

its a sequence convergence

Answer 10

Alright, then what you have looks good.

Answer 11

how did he get the upper bound, of 1/n

Answer 12

?? Ratio

\(\dfrac{\dfrac{(n+1)!}{(n+1)^{n+1}}}{\dfrac{n!}{n^{n}}} = \dfrac{(n+1)!}{n!}\dfrac{n^{n}}{(n+1)^{n+1}} = (n+1)\dfrac{n^{n}}{(n+1)^n}\dfrac{1}{n+1} = \left(\dfrac{n}{n+1}\right)^{n}\)

\(\lim\limits_{n\rightarrow\infty}\left(\dfrac{n}{n+1}\right)^{n} = \dfrac{1}{e} < 1\)

Answer 13

didnt really answer the question about the upper bound of 1/n, but u get points anway

Answer 14

ill ask it this way

Answer 15

how is (1/n)*(2/n)...n/n < 1/n

Answer 16

take for example n = 3. then 1*2*3/3 is not less than 1/3

Answer 17

For large enough n

2/n < 1
3/n < 1
,,,
n/n = 1

Thus: (2/n)(3/n)...(n/n) < 1

Thus (1/n)(2/n)(3/n)...(n/n) < 1/n

(1/1)(1/2)(1/3) < 1/3

Answer 18

why is your n changing, u said (1/n)(2/n)(3/n)...(n/n) < 1/n. but then the next line you give n different values.

Answer 19

Whoops. Good call

(1/n)(2/n)(3/n) < 3/n because (1/n)(2/n) < 1

But we can do better than that.

(1/n)(2/n)(3/n) < 1/n because (3/n)(2/n) < 1

Answer 20

ok, so to clarify. (1/n)(2/n)(3/n) < 1/n because (3/n)(2/n) < 1 because (3/3)*(2/3)<1

Answer 21

well because because (3/3)*(2/3)*(1/3)<1