Question

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

Answer 1

`-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`