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

Any help?

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- Sep 25th 2006, 02:30 PMJaysFan31Limit of Sequence
Prove that the limit as n approaches infinity of (2^n)/n! = 0.

Any help? - Sep 25th 2006, 04:57 PMPlato
Have you considered the limit of the ratio [a_(n+1)/a_n]->0?

- Sep 26th 2006, 08:44 AMclassicstrings
- Sep 26th 2006, 08:47 AMThePerfectHacker
- Sep 26th 2006, 08:51 AMclassicstrings
Maybe you can show me? Or not if you can't be bothered...:p

- Sep 26th 2006, 09:04 AMtopsquark
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, 09:35 AMThePerfectHacker
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, 10:28 AMPlato
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.