# Limit of Sequence

• Sep 25th 2006, 01:30 PM
JaysFan31
Limit of Sequence
Prove that the limit as n approaches infinity of (2^n)/n! = 0.

Any help?
• Sep 25th 2006, 03:57 PM
Plato
Have you considered the limit of the ratio [a_(n+1)/a_n]->0?
• Sep 26th 2006, 07:44 AM
classicstrings
Quote:

Originally Posted by JaysFan31
Prove that the limit as n approaches infinity of (2^n)/n! = 0.

Any help?

Well just by looking at it, 2^a number is less than the same number factorial for large values.

But that's no proof...more like logic...
• Sep 26th 2006, 07:47 AM
ThePerfectHacker
Quote:

Originally Posted by classicstrings
But that's no proof...more like logic...

I am sure he realized that.

He is probably taking a real analysis class he must post proofs. In calculus it is okay to say that but in a real analysis class you will fail if you say something like that.
• Sep 26th 2006, 07:51 AM
classicstrings
Maybe you can show me? Or not if you can't be bothered...:p
• Sep 26th 2006, 08:04 AM
topsquark
Quote:

Originally Posted by classicstrings
Maybe you can show me? Or not if you can't be bothered...:p

Plato's hint is probably the best way:

a_n = 2^n /n!

So lim(n->inf) a_(n+1)/a_n = lin(n->inf)[2^(n+1)/(n+1)!]/[2^n/n!]

= lim(n->inf)[2^(n+1)/2^n][n!/(n+1)!] = lim(n->inf)[2/(n+1)] = 0.

Since this limit goes to zero, the limit of a_n goes to zero as n->infinity.

-Dan
• Sep 26th 2006, 08:35 AM
ThePerfectHacker
Congratulations topsquark on your 1:):):)th post.
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The name of that rule is called the ratio test.
If you ever studied infinite series that is what is used.
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I have very elegant explanation to why it works (again if you studied infinite series).

You are given that sequence a_n
Instead consider the sequence of partial sums s_n
Since, by the ratio test lim s_n exists it implies that the originial sequence must satisfy lim a_n --> 0
• Sep 26th 2006, 09:28 AM
Plato
If {a_n+1/a_n}->0 and if 0<r<1 there is a positive integer K such that {a_K+1/a_K}<r.
That implies that {a_K+1}<{a_K}r.
But then {a_K+2/a_K+1}<r or {a_K+2}<{a_K+1}r<{a_K}r^2.
We get for each n, {a_K+n}r<{a_K}r^n.
Also because 0<r<1, (r^n)->0 so {a_K+n}->0.