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# 数学代写|微积分代写Calculus代考|MATH-172 Level comparison test examples

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## 数学代写|微积分代写Calculus代考|Level comparison test examples

Let’s begin by reprising example 4 .
Example 5 Determine the convergence or divergence of $\sum_{n=2}^{\infty} \frac{2}{n^2-1}$.
Solution To use the level comparison test, we calculate $\frac{1}{a_{\Omega}}$ and determine whether this quantity is in the convergence zone or the divergence zone (if we can tell). For any positive infinite hyperreal integer $\Omega$,
$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{2}{\Omega^2-1}}=\frac{1}{2}\left(\Omega^2-1\right) \approx \frac{1}{2} \Omega^2,$$
which is on the $\Omega^2$ level. Because the $\Omega^2$ level is in the convergence zone, the series converges.

Although the (direct) comparison test was inconclusive for the series of example 4, the level comparison test handled the series with ease. The same is true for the next example.
Example 6 Test $\sum_{n=4}^{\infty} \frac{n^2-3 n}{\sqrt[3]{n^{10}-4 n^2}}$ for convergence or divergence.
Solution Try the level comparison test. For any positive infinite hyperreal integer $\Omega$,
$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{\Omega^2-3 \Omega}{\sqrt[3]{\Omega^{10}-4 \Omega^2}}} \approx \frac{\sqrt[3]{\Omega^{10}}}{\Omega^2}=\frac{\Omega^{10 / 3}}{\Omega^2}=\Omega^{4 / 3} .$$
Because $\Omega^{4 / 3}$ is in the convergence zone ( $\Omega^p$ for $\left.p=\frac{4}{3}>1\right)$, the series converges.

## 数学代写|微积分代写Calculus代考|Locating a level in the correct zone

Next we reprise example 2, which we worked earlier using the comparison test.
Example 11 Determine the convergence or divergence of $\sum_{n=3}^{\infty} \frac{\ln n}{n}$.
Solution Try the level comparison test. For any positive infinite hyperreal integer $\Omega$,
$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{\ln \Omega}{\Omega}}=\frac{\Omega}{\ln \Omega} .$$
Although $\frac{\Omega}{\ln \Omega}$ is not listed in the diagram of figure 7 , because $\ln \Omega$ is infinite, we know that $\frac{\Omega}{\ln \Omega}$ is on a lower level than $\Omega$. Because $\Omega$ is in the divergence zone, so is $\frac{\Omega}{\ln \Omega}$, and the series diverges.

Regardless of whether the solution featured in example 11 is preferred over the solution featured in example 2 , the lesson to be learned is that we sometimes must be creative in determining the placement of a level in the convergence or divergence zone. Knowing the relative placement of a level is often enough.

## 数学代写|微积分代写Calculus代考|Level comparison test examples

$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{2}{\Omega^2-1}}=\frac{1}{2}\left(\Omega^2-1\right) \approx \frac{1}{2} \Omega^2,$$

$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{\Omega^2-3 \Omega}{\sqrt[3]{\Omega^{10}-4 \Omega^2}}} \approx \frac{\sqrt[3]{\Omega^{10}}}{\Omega^2}=\frac{\Omega^{10 / 3}}{\Omega^2}=\Omega^{4 / 3} .$$

## 数学代写|微积分代写Calculus代考|Locating a level in the correct zone

$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{\ln \Omega}{\Omega}}=\frac{\Omega}{\ln \Omega} .$$

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