# Use the limit Comparison Test to detemine the convergence or divergence of the series. sum_{n=1}^inftyfrac{5}{n+sqrt{n^2+4}}

Question
Series
Use the limit Comparison Test to detemine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{5}{n+\sqrt{n^2+4}}$$

2020-12-15
Given Data
The series is $$\sum_{n=1}^\infty\frac{5}{n+\sqrt{n^2+4}}$$
Let,
$$a_n=\frac{5}{n+\sqrt{n^2+4}}$$
$$b_n=\frac{1}{n}$$
Evaluate the convergence or divergence of the series by using limit ratio test,
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(\frac{5}{n+\sqrt{n^2+4}})}{(\frac{1}{n})}$$
$$=\lim_{n\to\infty}(\frac{5n}{n+\sqrt{n^2+4}})$$
$$=\lim_{n\to\infty}(\frac{5n}{n+n\sqrt{1+\frac{4}{n^2}}})$$
$$=\lim_{n\to\infty}(\frac{5n}{n(1+\sqrt{1+\frac{4}{n^2}})})$$
$$=\frac{5}{1+\sqrt{1+\frac{4}{\infty}}}$$
$$=\frac{5}{1+1}$$
$$=\frac{5}{2}$$
$$>0$$
The value of series by using the limit comparison test is $$\frac{5}{2}$$, which is greater than 0.
Hence the series is divergent.

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