Use the Limit Comparison Test to prove convergence or divergence of the infinite series.

$\sum _{n=2}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}$

UkusakazaL
2021-01-31
Answered

Use the Limit Comparison Test to prove convergence or divergence of the infinite series.

$\sum _{n=2}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}$

You can still ask an expert for help

Arnold Odonnell

Answered 2021-02-01
Author has **109** answers

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$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})=$ 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$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})=0$ and $\sum _{n\to \mathrm{\infty}}{b}_{n}$ converges, then series $\sum _{n\to \mathrm{\infty}}{a}_{n}$ will also converge.

3. If$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})\to \mathrm{\infty}$ 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$(\frac{{a}_{n}}{{b}_{n}})$ as,

$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})=\left(\frac{\frac{n}{\sqrt{{n}^{3}-1}}}{\frac{1}{\sqrt{n}}}\right)$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{n}{\sqrt{{n}^{3}-1}}\times \sqrt{n})$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{\sqrt{{n}^{3}}}{\sqrt{{n}^{3}-1}})$

$=\underset{n\to \mathrm{\infty}}{lim}(\sqrt{\frac{{n}^{3}}{{n}^{3}-1}})$

$=\underset{n\to \mathrm{\infty}}{lim}(\sqrt{\frac{1}{1-\frac{1}{{n}^{3}}}})$

As$n\to \mathrm{\infty}$ , then $\frac{1}{{n}^{3}}\to 0$ ,

$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})=\sqrt{\frac{1}{1-0}}$

$=1$

Since$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{b}_{n}})$ 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,

$\sum _{n=2}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}=\sum _{n=1}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}-[\frac{n}{\sqrt{{n}^{3}+1}}{]}_{n=1}$

$=\sum _{n=1}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}-\frac{1}{\sqrt{{1}^{3}+1}}$

$=\sum _{n=1}^{\mathrm{\infty}}\frac{n}{\sqrt{{n}^{3}+1}}-\frac{1}{\sqrt{2}}$

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.

Consider another function

According to the limit comparison test-

1. If

2. If

3. If

Clearly the series

Check fot he series

As

Since

Consider the following,

Consider that the difference between two finite values is also finite.

Thus, it is concluded that the series

Jeffrey Jordon

Answered 2021-12-27
Author has **2262** answers

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So my problem is that I have to use implicit differentiation to find the derivative of $F(x,y)={e}^{xy}-x$ when $F(x,y)=10$ and the equation of the tangent at the point $(1,\mathrm{log}(11))$. So I tried to solve this using two ways:

The first way I used was rearranging the equation to $10+x={e}^{xy}$ and then using $\mathrm{ln}$ to simply into $\mathrm{ln}(10+x)=xy$ before using implicit differentiation. The result is

$\frac{dy}{dx}=\frac{1}{10x+{x}^{2}}-\frac{y}{x}$

and after substituting the point, I get approximately −0.950 as the slope.

The second way I used was just differentiating it straight away rather than rearranging and I get

$\frac{dy}{dx}=\frac{\frac{1}{{e}^{xy}}-y}{x}.$

But when I sub the point in I get approximately −0.688 as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?

The first way I used was rearranging the equation to $10+x={e}^{xy}$ and then using $\mathrm{ln}$ to simply into $\mathrm{ln}(10+x)=xy$ before using implicit differentiation. The result is

$\frac{dy}{dx}=\frac{1}{10x+{x}^{2}}-\frac{y}{x}$

and after substituting the point, I get approximately −0.950 as the slope.

The second way I used was just differentiating it straight away rather than rearranging and I get

$\frac{dy}{dx}=\frac{\frac{1}{{e}^{xy}}-y}{x}.$

But when I sub the point in I get approximately −0.688 as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?