`sum_(n=1)^oo (2^n+1)/(5^n+1)` Use the Limit Comparison Test to determine the convergence or divergence of the series.

Expert Answers
gsarora17 eNotes educator| Certified Educator

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.