Does #a_n=n^n/(n!) # converge?

2 Answers
George C.
Nov 5, 2015

No, but #a_(n+1)/a_n -> e# as #n->oo#

Explanation:

#a_(n+1) = (n+1)^(n+1)/((n+1)!)#

#=(((n+1)/n)^n n^n (n+1))/(n!(n+1))#

#=((1+1/n)^n n^n)/(n!) = (1+1/n)^n a_n#

Now #lim_(n->oo) (1+1/n)^n = e#

So #lim_(n->oo) (a_(n+1)/a_n) = e#

Bill K.
Nov 5, 2015

In addition, the fact that #a_{n+1}/a_{n} -> e > 1# as #n->infty#, as shown already, also implies that the infinite series #sum_{n=1}^{infty}a_{n}=sum_{n=1}^{infty}(n^(n))/(n!)# diverges, by the Ratio Test .

Explanation:

The Ratio Test says that a series #sum_{n=1}^{infty}a_{n}# converges if #lim_{n->infty}|a_{n+1}|/|a_{n}|<1# and diverges if #lim_{n-> infty}|a_{n+1}|/|a_{n}|>1#, assuming these limits exist (or, in the second case, assuming the sequence #|a_{n+1}|/|a_{n}|# diverges to #infty#).

Related questions
See all questions in Infinite Sequences
Impact of this question
2342 views around the world
You can reuse this answer
Creative Commons License