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

Question

1594 views around the world

You can reuse this answer
Creative Commons License

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#