Question

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

Answer 1

The series:

`sum_(n=2)^oo (2^nx^n)/lnn`

is convergent for `x in [-1/2,1/2)` and absolutely convergent in the interior of the interval. The radius of convergence is `R=1/2`

Explanation:

Given:

`a_n = (2^nx^n)/lnn`

Evaluate the ratio:

`abs(a_(n+1)/a_n) = ( (2^(n+1)x^(n+1))/ln(n+1) ) / ( (2^nx^n)/lnn)`

`abs(a_(n+1)/a_n) = (2^(n+1)x^(n+1))/ (2^nx^n) (lnn ) /ln(n+1)`

`abs(a_(n+1)/a_n) = 2x (lnn ) /ln(n+1)`

The limit is then:

`lim_(n->oo) abs(a_(n+1)/a_n) = 2x lim_(n->oo) (lnn ) /ln(n+1) =2x`

so based on the ratio test we have that the series is absolutely convergent for `abs(2x) < 1` and divergent for `abs(2x) > 1`.

For `abs(2x) =1` the test is inconclusive and we must verify case by case:

A. For `x=1/2` we have:

`a_n = (2^nx^n)/lnn = 2^n(1/2)^n1/lnn = 1/lnn`

which can be proved to be divergent by direct comparison as `1/lnn > 1/n` and the harmonic series is divergent.

B. For `x=-1/2` we have:

`a_n = (2^nx^n)/lnn = 2^n(-1/2)^n1/lnn = (-1)^n/lnn`

which can be proved to be convergent based on Leibniz' theorem, as:

`lim_(n->oo) 1/lnn = 0`

and

`1/lnn > 1/ln(n+1)`

In conclusion the series:

`sum_(n=2)^oo (2^nx^n)/lnn`

is convergent for `x in [-1/2,1/2)` and absolutely convergent in the interior of the interval.