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

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

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Lacey-May Snyder

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 $$4^n+1>4^n$$
$$\Rightarrow\frac{1}{4^n+1}<\frac{1}{4^n}$$
$$\Rightarrow\frac{5}{4^n+1}<\frac{5}{4^n}$$
so, $$0<a_n<\frac{5}{4^n}$$
Thus, we can compare the given series $$\sum_{n=1}^\infty\frac{5}{4^n+1}$$ 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 $$|\frac14|<1$$
so the comparison test tells us that $$\sum_{n=1}^\infty\frac{5}{4^n+1}$$ also converges.

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