sum_(n=1)^oo n/((n+1)2^(n-1)) Use the Limit Comparison Test to determine the convergence or divergence of the series.

Limit comparison test is applicable when suma_n and sumb_n are series with positive terms . If lim_(n->oo)a_n/b_n=L where L is a finite number and L>0 , then either both series converge or both diverge.
Given series is sum_(n=1)^oon/((n+1)2^(n-1))
Let the comparison series be sum_(n=1)^oo1/2^n=sum_(n=1)^oo(1/2)^n
The comparison series is a geometric series with ratio r=1/2<1
A geometric series with ratio r converges, if 0<|r|<1
So, the comparison series sum_(n=1)^oo(1/2)^n converges.
Now ,let's apply the limit comparison test with a_n=n/((n+1)2^(n-1)) and b_n=1/2^n
a_n/b_n=(n/((n+1)2^(n-1)))/(1/2^n)
a_n/b_n=((2n)/((n+1)2^n))/(1/2^n)
a_n/b_n=(2n*2^n)/((n+1)2^n)
a_n/b_n=(2n)/(n+1)
lim_(n->oo)a_n/b_n=lim_(n->oo)(2n)/(n+1)
=lim_(n->oo)(2n)/(n(1+1/n))
=lim_(n->oo)2/(1+1/n)
=2>0
Since the comparison series sum_(n=1)^oo1/2^n converges, so the series sum_(n=1)^oon/((n+1)2^(n-1)) as well ,converges by the limit comparison test.