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

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

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Answer 1 · 2017-04-04T02:59:11

$-sqrt5lt=xltsqrt5$

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:

$-sqrt5lt=xltsqrt5$

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

1 Answer

Explanation:

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

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