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

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Answer 1 · 2017-02-06T17:53:24

converges for #x in RR^+#

Using the Stirling asymptotic approximation

#n! approx (n/e)^n# we have

#x^n/((x cdot n)!) approx x^n/((x cdot n)/e)^(x cdot n)=(x/x^x(e/n)^x)^n#

The sequence converges if #x/x^x(e/n)^x<1# so

#e/n < x^((x-1)/x)# Now given a #n# such that #e/n le epsilon#

then the convergence is attained for #x^((x-1)/x) > epsilon#

The function #f(x)=x^((x-1)/x) # for #x in RR^+# has a minimum at #x=1# and #f(1)=1#. Concluding the sequence is convergent for #x in RR^+#

Attached a plot of #f(x)# for #x in RR^+#

enter image source here

Answer 2 · 2017-04-19T12:24:48

Yes, #lim_(n to infty)a_n=0# for all real numbers #x#.

Choose any real number #x#, and let #N# be any natural number greater than or equal to #|x|#.

Let us look at #|a_n|#.

#|a_n|=|x^n/(x n!)|=|x|^(n-1)/(n!) leq (N^(n-1))/(n!)#

#=1/ncdot (N/(n-1)cdot N/(n-2)cdots N/(N+1)) cdot (N/N cdot N/(N-1)cdotsN/1)#

#leq 1/n cdot (N/N cdot N/NcdotsN/N) cdot (N^N/(N!))=1/n(N^N/(N!))#

So, we have

#Rightarrow 0 leq |a_n| leq 1/n(N^N/(N!)) to 0# as #n to 0#

By Squeeze Theorem,

#Rightarrow lim_(n to infty)|a_n|=0#

which implies

#Rightarrow lim_(n to infty)a_n=0#

I hope that this was clear.