How do you determine whether the sequence #a_n=(n!+2)/((n+1)!+1)# converges, if so how do you find the limit?

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Answer 1 · 2017-06-18T22:35:15

The sequence #a_n# converges to #0#. The series (i.e. sum) diverges.

Let:

#a_n = (n!+2)/((n+1)!+1)#

#color(white)(a_n) = (n!+2)/((n+1)n!+1)#

#color(white)(a_n) = (1+2/(n!))/(n+1+1/(n!))#

As #n->oo#:

#2/(n!)->0#

#1/(n!)->0#

So:

#a_n->(1+0)/(oo+1+0) = 0#

So the sequence #a_1, a_2, a_3,...# tends to #0# as #n->oo#

How about the sum?

#a_n = (n!+2)/((n+1)!+1) > (n!)/((n+1)!+1) = 1/(n+1+1/(n!)) >= 1/(n+2)#

So:

#sum_(n=1)^oo a_n >= sum_(n=1)^oo 1/(n+2) = sum_(n=3)^oo 1/n#

which diverges, since the harmonic series #sum_(n=1)^oo 1/n# diverges.