# `sum_(n=1)^oo 5/(4^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)^oo5/(4^n+1)`

Let the comparison series be `sum_(n=1)^oo1/4^n=sum_(n=1)^oo(1/4)^n`

The comparison series `sum_(n=1)^oo(1/4)^n` is a geometric series with `r=1/4<1`

A geometric series with ratio r converges if `0<|r|<1`

So, the comparison series which is a geometric series converges.

Now let's use the limit comparison test with:

`a_n=5/(4^n+1)`   and `b_n=1/4^n`

`a_n/b_n=(5/(4^n+1))/(1/4^n)`

`a_n/b^n=(5*4^n)/(4^n+1)`

`a_n/b_n=(5*4^n)/(4^n(1+1/4^n))`

`a_n/b_n=5/(1+1/4^n)`

`lim_(n->oo)a_n/b_n=lim_(n->oo)5/(1+1/4^n)`

`=5>0`

Since the comparison series `sum_(n=1)^oo(1/4)^n` converges, so the series `sum_(n=1)^oo5/(4^n+1)`  as well ,converges as per the limit comparison test.

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