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

Question
Series
asked 2021-01-31
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\(\sum_{n=2}^\infty\frac{n}{\sqrt{n^3+1}}\)

Answers (1)

2021-02-01
Let the given expression be represented as,
\(a_n=\frac{n}{\sqrt{n^3+1}}\)
Consider another function \(b_n=\frac{1}{\sqrt{n}}\)
According to the limit comparison test-
1. If \(\lim_{n\rightarrow\infty}(\frac{a_n}{b_n})=\) finite positive value, then series \(\sum_{n\to\infty}a_n\) and \(\sum_{n\to\infty}b_n\) behave alike, i.e. either both converge or both diverge.
2. If \(\lim_{n\rightarrow\infty}(\frac{a_n}{b_n})=0\) and \(\sum_{n\to\infty}b_n\) converges, then series \(\sum_{n\to\infty}a_n\) will also converge.
3. If \(\lim_{n\rightarrow\infty}(\frac{a_n}{b_n})\to\infty\) diverges, then series \(\sum_{n\to\infty}a_n\) will also diverge.
Clearly the series \(\sum_{n=1}^\infty b_n\) is a converging series as the term \(\lim_{n\to\infty}\frac{1}{\sqrt n}\approx0\)
Check fot he series \((\frac{a_n}{b_n})\) as,
\(\lim_{n\rightarrow\infty}(\frac{a_n}{b_n})=\left(\frac{\frac{n}{\sqrt{n^3-1}}}{\frac{1}{\sqrt{n}}}\right)\)
\(=\lim_{n\to\infty}(\frac{n}{\sqrt{n^3-1}}\times\sqrt{n})\)
\(=\lim_{n\to\infty}(\frac{\sqrt{n^3}}{\sqrt{n^3-1}})\)
\(=\lim_{n\to\infty}(\sqrt{\frac{n^3}{n^3-1}})\)
\(=\lim_{n\to\infty}(\sqrt{\frac{1}{1-\frac{1}{n^3}}})\)
As \(n\to\infty\), then \(\frac{1}{n^3}\to0\),
\(\lim_{n\to\infty}(\frac{a_n}{b_n})=\sqrt{\frac{1}{1-0}}\)
\(=1\)
Since \(\lim_{n\to\infty}(\frac{a_n}{b_n})\) is a finite positive value and \(\sum_{n=1}^\infty b_n\) is converging, therefore the limit comparison test implies that, the series \(\sum_{n=1}^\infty a_n\) is also converging.
Consider the following,
\(\sum_{n=2}^\infty\frac{n}{\sqrt{n^3+1}}=\sum_{n=1}^\infty\frac{n}{\sqrt{n^3+1}}-[\frac{n}{\sqrt{n^3+1}}]_{n=1}\)
\(=\sum_{n=1}^\infty\frac{n}{\sqrt{n^3+1}}-\frac{1}{\sqrt{1^3+1}}\)
\(=\sum_{n=1}^\infty\frac{n}{\sqrt{n^3+1}}-\frac{1}{\sqrt{2}}\)
Consider that the difference between two finite values is also finite.
Thus, it is concluded that the series \(=\sum_{n=2}^\infty\frac{n}{\sqrt{n^3+1}}\) is converging.
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