Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞54n+1

Given series, $\sum _{n=1}^{\mathrm{\infty }}\frac{5}{4^n+1}$
we have to determine the convergence or divergence of the given series.
Consider $a_n=\frac{5}{4^n+1}$
Since $4^n+1>4^n$
$\Rightarrow \frac{1}{4^n+1}<\frac{1}{4^n}$
$\Rightarrow \frac{5}{4^n+1}<\frac{5}{4^n}$
so, $0<a_n<\frac{5}{4^n}$
Thus, we can compare the given series $\sum _{n=1}^{\mathrm{\infty }}\frac{5}{4^n+1}$ with the geometric series
$\sum _{n=1}^{\mathrm{\infty }}\frac{5}{4^n}=5\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4^n}$
this geometric series converges since $|\frac{1}{4}|<1$
so the comparison test tells us that $\sum _{n=1}^{\mathrm{\infty }}\frac{5}{4^n+1}$ also converges.

Answer is given below (on video)