Does the series #sum_(n=1)^oo((2n)/(n!))# converge or diverge?

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Answer 1 · 2017-11-15T07:53:35

The series converge

Let #a_n=(2n)/(n!)#

Perform the series ratio test

#|a_(n+1)/a_n|=|(2(n+1))/((n+1)!)*((n!)/(2n))|#

#=|((n+1)*(n!))/((n+1)!n)|#

#=1/n# as #n in NN#

#lim_(n->o)|a_(n+1)/a_n|=lim_(n->o)(1/n)=0#

As the limit is #<1#, the series converge

Answer 2 · 2017-11-15T09:01:58

See below.

Recall that #e^x=sum_(n=0)^oox^n/(n!)#

Then, #sum_(n=1)^oo (2n)/(n!) =2sum_(n=1)^oo 1/((n-1)!) =2sum_(n=0)^oo 1/(n!)=2e#

so it converges.