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

1 Answer
Andrea S.
Oct 30, 2017

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.

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