Question

Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{1}{sqrt{n^3+2n}}

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asked 2020-12-28
Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}\)

Answers (1)

2020-12-29
Limit comparison test:
Suppose that we have two series \(\sum a_n\) and \(\sum b_n\), define,
\(b_n=\frac{1}{n^{\frac{3}{2}}}\)
Where L is positive and finite then either both series are convergent or both series divergent.
Given that,
\(\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}\)
Since we have,
\(\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}\)
Here
\(a_n=\frac{1}{\sqrt{n^3+2n}}\)
So,
\(b_n=\frac{1}{n^{\frac{3}{2}}}\)
Now evaluate \(b_n=\frac{1}{n^{\frac{3}{2}}}\)
\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^3+2n}}}{\frac{1}{n^{\frac{3}{2}}}}\)
\(=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^3+2n}}\times n^{\frac{3}{2}}\right)\)
\(=\lim_{n\to\infty}\left(\frac{1}{n^{\frac{3}{2}}\sqrt{1+\frac{2}{n^2}}}\times n^{\frac{3}{2}}\right)\)
\(=\lim_{n\to\infty}\left(\frac{1}{\sqrt{1+\frac{2}{n^2}}}\right)\)
\(=1\)
Since \(\lim_{n\to\infty}\frac{a_n}{b_n}=L=1\) which is finite and positive.
Since using p-series test [\(\sum\frac{1}{n^p}\) is convergent if \(p>1\) or divergent if \(p\leq1\)], \(b_n=\frac{1}{n^{\frac{3}{2}}}\) is convergent.
Since \(\lim_{n\to\infty}\frac{a_n}{b_n}\) is finite and positive and \(\sum b_n\) is convergent so using limit comparison test \(\sum a_n\) is also convergent.
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