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
asked 2020-12-14
Use the limit Comparison Test to detemine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{5}{n+\sqrt{n^2+4}}\)

Answers (1)

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.
0

Relevant Questions

asked 2020-11-24
Use the limit Comparison Test to detemine the convergence or divergence of the series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{5}}}{{{n}+\sqrt{{{n}^{{2}}+{4}}}}}}\)
asked 2020-12-16
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{5}{4^n+1}\)
asked 2020-12-28
Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}\)
asked 2020-11-29
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\(\sum_{n=1}^\infty\frac{1}{\sqrt n+\ln n}\)
asked 2021-01-31
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\(\sum_{n=2}^\infty\frac{n}{\sqrt{n^3+1}}\)
asked 2021-02-06
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{n^{k-1}}{n^k+1},k>2\)
asked 2021-01-10
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{e}^{{n}}+{n}}}{{{e}^{{{2}{n}}}-{n}^{{2}}}}}\)
asked 2020-11-08
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}^{{{k}-{1}}}}}{{{n}^{{k}}+{1}}}},{k}{>}{2}\)
asked 2021-01-28
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{n}{(n+1)2^{n-1}}\)
asked 2021-03-08
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}\)
...