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

Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
$\sum _{n=2}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}$
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Arnold Odonnell
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 $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)=$ finite positive value, then series $\sum _{n\to \mathrm{\infty }}{a}_{n}$ and $\sum _{n\to \mathrm{\infty }}{b}_{n}$ behave alike, i.e. either both converge or both diverge.
2. If $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)=0$ and $\sum _{n\to \mathrm{\infty }}{b}_{n}$ converges, then series $\sum _{n\to \mathrm{\infty }}{a}_{n}$ will also converge.
3. If $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)\to \mathrm{\infty }$ diverges, then series $\sum _{n\to \mathrm{\infty }}{a}_{n}$ will also diverge.
Clearly the series $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ is a converging series as the term $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n}}\approx 0$
Check fot he series $\left(\frac{{a}_{n}}{{b}_{n}}\right)$ as,
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)=\left(\frac{\frac{n}{\sqrt{{n}^{3}-1}}}{\frac{1}{\sqrt{n}}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{n}{\sqrt{{n}^{3}-1}}×\sqrt{n}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\sqrt{{n}^{3}}}{\sqrt{{n}^{3}-1}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sqrt{\frac{{n}^{3}}{{n}^{3}-1}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\sqrt{\frac{1}{1-\frac{1}{{n}^{3}}}}\right)$
As $n\to \mathrm{\infty }$, then $\frac{1}{{n}^{3}}\to 0$,
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)=\sqrt{\frac{1}{1-0}}$
$=1$
Since $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{b}_{n}}\right)$ is a finite positive value and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ is converging, therefore the limit comparison test implies that, the series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ is also converging.
Consider the following,
$\sum _{n=2}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}-\left[\frac{n}{\sqrt{{n}^{3}+1}}{\right]}_{n=1}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}-\frac{1}{\sqrt{{1}^{3}+1}}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}-\frac{1}{\sqrt{2}}$
Consider that the difference between two finite values is also finite.
Thus, it is concluded that the series $=\sum _{n=2}^{\mathrm{\infty }}\frac{n}{\sqrt{{n}^{3}+1}}$ is converging.
Jeffrey Jordon