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# 数学代写|微积分代写Calculus代考|Math323 Absolute and conditional convergence

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## 数学代写|微积分代写Calculus代考|Absolute and conditional convergence

We know that the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges; the sum of its terms is infinite. But, by alternating the signs of its terms, we can gain convergence; $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges by the alternating series test. The movement to the right (the positive terms) and the movement to the left (the negative terms), each of which totals an infinite distance, is balanced enough that the sum is finite.

By contrast, take a series such as $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a $p$-series with $p=2>1$ and therefore converges. The sum of its terms is finite. If we then alternate the signs of its terms to form $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$, which also converges by the alternating series test, convergence seems manifest from the outset. The total movement to the right and the total movement to the left are each finite and no balancing act is necessary.
Each of these alternating series converges, but somehow the convergence doesn’t feel the same. The convergence of the alternating series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ seems more fundamental, whereas the convergence of the alternating series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ seems much more fragile. These two situations are given names in the following definition.

## 数学代写|微积分代写Calculus代考|Absolute and conditional convergence: examples

We now have not just two but three options to choose from when testing a series.

Example 1 Does $\sum_{n=1}^{\infty} \frac{(-2)^{n+1}}{n+5^n}$ converge absolutely, converge conditionally, or diverge?

Solution Following the decision tree of figure 1, we first test for absolute convergence.
(1) Does $\sum_{n=1}^{\infty}\left|a_n\right|$ converge? We test the series $\sum_{n=1}^{\infty} \frac{2^{n+1}}{n+5^n}$. Applying the level comparison test, for any positive infinite hyperreal integer $\Omega$,
$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{2^{\Omega+1}}{\Omega+5^{\Omega}}}=\frac{\Omega+5^{\Omega}}{2^{\Omega+1}} \approx \frac{5^{\Omega}}{2^{\Omega+1}}=\frac{5^{\Omega}}{2 \cdot 2^{\Omega}}=\frac{1}{2}\left(\frac{5}{2}\right)^{\Omega},$$

which is on the $\left(\frac{5}{2}\right)^{\Omega}$ level and therefore in the convergence zone. Because $\sum_{n=1}^{\infty}\left|a_n\right|$ converges, the series $\sum_{n=1}^{\infty} \frac{(-2)^{n+1}}{n+5^n}$ converges absolutely.
A different path through the decision tree is illustrated next.

## 数学代写|微积分代写Calculus代考|Absolute and conditional convergence:examples

$$\frac{1}{a_{\Omega}}=\frac{1}{\frac{2^{\Omega+1}}{\Omega+5^{\Omega}}}=\frac{\Omega+5^{\Omega}}{2^{\Omega+1}} \approx \frac{5^{\Omega}}{2^{\Omega+1}}=\frac{5^{\Omega}}{2 \cdot 2^{\Omega}}=\frac{1}{2}\left(\frac{5}{2}\right)^{\Omega},$$

## MATLAB代写

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