Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{5}{4^n+1}

Question
Series
asked 2020-12-16
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{5}{4^n+1}\)

Answers (1)

2020-12-17
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.
0

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