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

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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