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

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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`









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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