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`
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.