# `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.

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