sum_(n=1)^oo (2^n+1)/(5^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)^oo(2^n+1)/(5^n+1)
Let the comparison series be sum_(n=1)^oo2^n/5^n=sum_(n=1)^oo(2/5)^n
The comparison series sum_(n=1)^oo(2/5)^n is a geometric series with ratio r=2/5<1
A geometric series 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=(2^n+1)/(5^n+1)   and b_n=2^n/5^n
a_n/b_n=((2^n+1)/(5^n+1))/(2^n/5^n)
a_n/b^n=(2^n+1)/(5^n+1)(5^n/2^n)
a_n/b^n=((2^n+1)/2^n)(5^n/(5^n+1))
a_n/b^n=(1+1/2^n)(1/(1+1/5^n))
lim_(n->oo)a_n/b_n=lim_(n->oo)(1+1/2^n)(1/(1+1/5^n))
=1>0
Since the comparison series sum_(n=1)^oo2^n/5^n converges,the series sum_(n=1)^oo(2^n+1)/(5^n+1) as well ,converges as per the limit comparison test.