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