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

Question

14569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-03T00:04:28

#R=1/e#

Use the ratio test, which states that #suma_n# converges if #L<1#, where #L=lim_(nrarroo)abs(a_(n+1)/a_n)#. Here,

#L=lim_(nrarroo)abs((n+1)^(n+1)/((n+1)!)x^(n+1)(n!)/n^n1/x^n)#

Simplifying:

#L=lim_(nrarroo)abs((n+1)^(n+1)/n^n((n!)/((n+1)n!))x^(n+1)/x^n)#

#L=lim_(nrarroo)abs((n+1)^(n+1)/(n+1)1/n^nx)#

Bringing the #x# out of the limit, since the limit depends only on how #n# changes:

#L=absxlim_(nrarroo)abs((n+1)^n/n^n)#

#L=absxlim_(nrarroo)abs(((n+1)/n)^n)#

This is a well-known limit that approaches #e#:

#L=eabsx#

The posted series will converge when #L<1#, so the interval of convergence will be on when:

#eabsx<1#

#absx<1/e#

Thus the radius of convergence is #R=1/e#, centered at #x=0#.