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

Question

Elijah Abram · Accepted Answer

The sequence #a_n->0# for any fixed #x# For any fixed #x = x_0# there is a #n_0# such that for #n>n_0# the sequence #a_n->0#. The converse is easy to obtain. Given a #n_0# then fixing a #x_0 = log_e(n_0!)/log_en_0#, for #n>n_0# then #a_n(x_0)->0#

Elijah Abram · Answer

For every #x#, #lim_(nrarroo)n^x/(n!) = 0#

Exponential grows faster than power.

#AAx# #lim_(nrarroo)n^x/(e^n) = 0#

Factorial grows faster than exponential.

#lim_(nrarroo)e^n/(n!) = 0#

The product must also go to #0#

Ezekiel Alexander · Answer

undefined The sequence $ a_n = \frac{n^x}{n!} $ converges if and only if the limit of $ a_n $ as $ n $ approaches infinity exists and is finite. This convergence is determined by the value of $ x $.

For any fixed value of $ x $, the sequence $ a_n $ converges if $ x $ is less than or equal to zero. When $ x $ is a non-negative integer or zero, the sequence converges to $ 0 $ as $ n $ approaches infinity.

However, if $ x $ is a positive non-integer, the sequence $ a_n $ diverges as $ n $ approaches infinity. This is because the factorial $ n! $ grows faster than any power of $ n $, causing the terms $ \frac{n^x}{n!} $ to approach zero too slowly for convergence.

Therefore, the sequence $ a_n = \frac{n^x}{n!} $ converges only for $ x \leq 0 $ and diverges for $ x > 0 $.