Given series, \(\sum_{n=1}^\infty\frac{5}{4^n+1}\)

we have to determine the convergence or divergence of the given series.

Consider \(a_n=\frac{5}{4^n+1}\)

Since \(4^n+1>4^n\)

\(\Rightarrow\frac{1}{4^n+1}<\frac{1}{4^n}\)</span>

\(\Rightarrow\frac{5}{4^n+1}<\frac{5}{4^n}\)</span>

so, \(0

Thus, we can compare the given series \(\sum_{n=1}^\infty\frac{5}{4^n+1}\) with the geometric series

\(\sum_{n=1}^\infty\frac{5}{4^n}=5\sum_{n=1}^\infty\frac{1}{4^n}\)

this geometric series converges since \(|\frac14|<1\)</span>

so the comparison test tells us that \(\sum_{n=1}^\infty\frac{5}{4^n+1}\) also converges.

we have to determine the convergence or divergence of the given series.

Consider \(a_n=\frac{5}{4^n+1}\)

Since \(4^n+1>4^n\)

\(\Rightarrow\frac{1}{4^n+1}<\frac{1}{4^n}\)</span>

\(\Rightarrow\frac{5}{4^n+1}<\frac{5}{4^n}\)</span>

so, \(0

Thus, we can compare the given series \(\sum_{n=1}^\infty\frac{5}{4^n+1}\) with the geometric series

\(\sum_{n=1}^\infty\frac{5}{4^n}=5\sum_{n=1}^\infty\frac{1}{4^n}\)

this geometric series converges since \(|\frac14|<1\)</span>

so the comparison test tells us that \(\sum_{n=1}^\infty\frac{5}{4^n+1}\) also converges.