How do you use the limit comparison test to determine if #Sigma n/((n+1)2^(n-1))# from #[1,oo)# is convergent or divergent?

1 Answer
Nov 20, 2016

#sum_(n=1)^oon/((n+1)2^(n-1))#

Rewriting this:

#=sum_(n=1)^oon/(n+1)1/(2^(n-1))=sum_(n=1)^oon/(n+1)1/(2^n/2)=sum_(n=1)^oon/(n+1)2/2^n#

Bringing the #2# out and noticing that #1/2^n=(1/2)^n#:

#=2sum_(n=1)^oon/(n+1)(1/2)^n#

We should recognize that #sum_(n=1)^oo(1/2)^n# is a geometric series, and since #1/2<1#, we know this series will converge.

Also note that when #n>0#, all terms of #n/(n+1)<1#. This means we can say that:

#n/(n+1)(1/2)^n<=(1/2)^n#

We can now use the direct comparison test. Since #sum_(n=1)^oo(1/2)^n# converges, and #n/(n+1)(1/2)^n<=(1/2)^n#, we know that #sum_(n=1)^oon/(n+1)(1/2)^n# converges as well.

Thus #2sum_(n=1)^oon/(n+1)(1/2)^n=sum_(n=1)^oon/((n+1)2^(n-1))# is convergent.