How do you find the radius of convergence $Sigma (x^n)/(5^(nsqrtn))$ from $n=[0,oo)$ ?

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Answer 1 · 2017-07-31T14:07:17

The radius of convergence for the series:

$sum_(n=0)^oo x^n/(5^(nsqrtn))$

is $R=5$ .

Evaluate the ratio:

$abs( a_(n+1)/a_n) = abs ( ( x^(n+1)/(5^((n+1)sqrt(n+1))))/(x^n/(5^(nsqrtn))))$

$abs( a_(n+1)/a_n) = abs ( x^(n+1)/x^n)5^(nsqrtn)/(5^((n+1)sqrt(n+1))$

$abs( a_(n+1)/a_n) = abs ( x)5^n/5^(n+1)(5^sqrtn)/(5^(sqrt(n+1))$

$abs( a_(n+1)/a_n) = abs ( x)/5 5^(sqrt(n)-sqrt(n+1))$

Now note that:

$lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) (sqrt(n)-sqrt(n+1)) xx (sqrt(n)+sqrt(n+1))/ (sqrt(n)+sqrt(n+1))$

$lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) (n-(n+1))/ (sqrt(n)+sqrt(n+1))$

$lim_(n->oo) (sqrt(n)-sqrt(n+1)) = lim_(n->oo) -1/ (sqrt(n)+sqrt(n+1)) = 0$

so that:

$lim_(n->oo) abs( a_(n+1)/a_n) = abs ( x)/5 lim_(n->oo)5^(sqrt(n)-sqrt(n+1)) = abs ( x)/5 xx 5^0 = abs ( x)/5$

Based on the ratio test the series is therefore absolutely convergent for:

$abs ( x)/5 < 1$

and not convergent for $abs ( x)/5 > 1$

So the radius of convergence is $R=5$

How do you find the radius of convergence Sigma (x^n)/(5^(nsqrtn))Sigma (x^n)/(5^(nsqrtn)) from n=[0,oo)n=[0,oo)?