Question

Show that \(\dfrac{3^n}{n!}\) converges to 0
Please help.

Answer 1

@dan815

Answer 2

\[\frac{3}{n}\frac{3}{n-1}\cdots\frac{3}{3}\frac{3}{2}\frac{3}{1}=\frac{3}{n}\times\frac{3}{n-1}\cdots\frac{3}{3}\times\frac{3}{2}\frac{3}{1}\le\frac{27}{2n}\]

Answer 3

note \[\frac{3}{n-1}\cdots\frac{3}{3}\le 1\]

Answer 4

Oh, You try to put in sandwich theorem, right?

Answer 5

yes...\[0\le\frac{3^n}{n!}\le\frac{27}{2n}\]

Answer 6

the far right is = (27/2) (1/n) which goes to 0 when n goes to infinitive, right?

Answer 7

yes

Answer 8

Is there any logic between the line? That is when we assume that the middle term has limit, it will be 0 .

Answer 9

by the squeeze theorem it has to be zero

Answer 10

it is trapped between to things that are going to zero

Answer 11

I lost my credits because of this mistake many times, but still not know how to fix it. My Prof commented : "How can you know whether the limit of the term exist or not?"

Answer 12

it is given in the theorem...
http://en.wikipedia.org/wiki/Squeeze_theorem#Statement

Answer 13

He said : "If it exists, then it is 0, but we don't know whether it exists or not yet"

Answer 14

squeeze theorem tells us it exists

Answer 15

and it has to have the same value as the other limits

Answer 16

Thank you. I review for final and I do not want to make mistakes. :)

Answer 17

@Zarkon This is the way how I lost my credit :(

Answer 18

ic what you did

Answer 19

state the inequalities without limits like I did above...

then find the limits of the left and right function....


finally (if the limits are the same) state by the squeeze theorem that the limit exists and is equal to the previous limits

Answer 20

understand?

Answer 21

Yes, Sir. Thank you. :)

Answer 22

harsh marker a complete 0 :P