How do you find the radius of convergence #Sigma x^n/lnn# from #n=[2,oo)#?

1 Answer
Andrea S.
Jan 24, 2017

The series:

#sum_(n=2)^oo x^n/lnn#

has radius of convergence #R=1#

Explanation:

To find the radius of convergence we can apply the ratio test, stating that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that:

#L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1#

If #L < 1 # the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

#abs (a_(n+1)/a_n) = abs ( frac ( x^(n+1) /ln(n+1)) ( x^n/lnn)) = abs(x) ln n/ln (n+1)#

Using the properties of logarithms:

#ln n = ln (n+1-1) = ln((n+1)(1-1/(n+1))) = ln(n+1) + ln(1-1/(n+1)) #

so:

#lim_(n->oo) ln n/ln (n+1) = lim_(n->oo) 1+ ln(1-1/(n+1))/ln (n+1) = 1#

and then:

#lim_(n->oo) abs(x)ln n/ln (n+1) = abs(x)#

so that the series is absolutely convergent for #abs(x) < 1#

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