Ow do you test for convergence for ∑ ln ( n )/ 2^n for n=1 to infinity?

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#2^n#

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Answer 1 · 2018-05-25T17:15:14

The series converges.

Let #a_n=ln(n)/2^n#

Then, calculate #lim_(n->oo)|a_(n+1)/a_n|#

So,

#|a_(n+1)/a_n|=|(ln(n+1)/2^(n+1))/((ln(n))/2^n)|#

#=|ln(n+1)/(2ln(n))|#

Then, as #ln(n+1)/(2ln(n))# is positive as #n->oo#

#lim_(n->oo)|ln(n+1)/(2ln(n))|=1/2lim_(n->oo)(ln(n+1)/(ln(n)))#

Apply L'Hopital's rule

#=1/2lim_(n->oo)(1/(n+1))/(1/n)#

#=1/2lim_(n->oo)(n/(n+1))#

Apply L'Hopital's rule

#=1/2lim_(n->oo)(1/(1))#

#=1/2#

As the limit by the ratio test is #<1#, the series converges