How do you find the interval of convergence #Sigma x^n/(6sqrtn)# from #n=[1,oo)#?
1 Answer
Explanation:
The series
#L=lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs((x^(n+1)/(6sqrt(n+1)))/(x^n/(6sqrtn)))=lim_(nrarroo)abs(x^(n+1)/x^n((6sqrtn)/(6sqrt(n+1)))#
We can simplify and move the
#L=absxlim_(nrarroo)abssqrt(n/(n+1))#
The limit approaches
#L=absx#
The series converges when
#absx<1#
#-1ltxlt1#
Before we say we're done, we have to check the endpoints
At
#sum_(n=1)^oo1^n/(6sqrtn)=1/6sum_(n=1)^oo1/sqrtn#
This is divergent through the p-series test, since
At
#1/6sum_(n=1)^oo(-1)^n/sqrtn#
Through the alternating series test, we see that this does converge, so
The interval is then:
#-1lt=xlt1#