Question

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

Answer 1

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`