`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

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial