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

Related questions
See all questions in Determining the Radius and Interval of Convergence for a Power Series
Impact of this question
15083 views around the world
You can reuse this answer
Creative Commons License