How do you find the interval of convergence #Sigma x^(2n)/(n*5^n)# from #n=[1,oo)#?

#-sqrt5lt=xltsqrt5#

Explanation:

Find the ratio #abs(a_(n+1)/a_n)# for the series #suma_n#:

#abs(a_(n+1)/a_n)=abs(x^(2(n+1))/((n+1)5^(n+1))((n(5^n))/x^(2n)))=abs(x^2/5(n/(n+1)))#

The series converges if #lim_(nrarroo)abs(a_(n+1)/a_n)<1#.

#lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(x^2/5(n/(n+1)))=x^2/5lim_(nrarroo)abs(n/(n+1))=x^2/5#

So the series is convergent for

#x^2/5<1#

#x^2-5<0#

#(x-sqrt5)(x+sqrt5)<0#

#-sqrt5 < x < sqrt5#

Test both these endpoints by plugging them into the original series expression. Note that #x=sqrt5# gives the divergent series #sum_(n=1)^oo 1/n#, whereas #x=-sqrt5# gives the conditionally convergent series #sum_(n=1)^oo(-1)^n/n#. Thus, #x=-sqrt5# is included in the interval:

#-sqrt5lt=xltsqrt5#