How do you find the radius of convergence of the power series #Sigma x^n/(n!)^(1/n)# from #n=[1,oo)#?

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Answer 1 · 2017-02-01T09:10:32

The series:

#sum_(n=1)^oo x^n/(n!)^(1/n)#

has radius of convergence #R=1#

We can apply the ratio test:

#abs(a_(n+1)/a_n) = (x^(n+1)/((n+1)!)^(1/n))/(x^n/(n!)^(1/n)) = abs(x^(n+1)/x^n (n!)^(1/n) / ((n+1)!)^(1/n)) = abs(x)/(n+1)^(1/n)#

now consider:

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

As #e^x# is continuous:

# lim_(n->oo) (n+1)^(1/n) = e^(lim_(n->oo) (ln(n+1)/n)) = e^0 = 1#

So we have:

#lim_(n->oo) abs(a_(n+1)/a_n) = abs(x)#

which means that the series is absolutely convergent for #abs(x)<1# and divergent for #abs(x) > 1#, that is the radius of convergence is #R=1#