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.