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

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Answer 1 · 2017-10-30T22:16:27

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#

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.