Thread: Limit of Sequence

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

Any help?

Have you considered the limit of the ratio [a_(n+1)/a_n]->0?

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...

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.

Maybe you can show me? Or not if you can't be bothered...

Originally Posted by classicstrings

Maybe you can show me? Or not if you can't be bothered...

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

Congratulations topsquark on your 1th 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

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.