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# Calculus_微积分_MAST10006 Some Tests for Convergence

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## Calculus_微积分_Some Tests for Convergence

Theorem 1.2.6 (Comparison test) Let $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}$. Then
$$\sum_{n=1}^{\infty} b_n \text { converges } \Rightarrow \sum_{n=1}^{\infty} a_n \text { converges. }$$
Proof Suppose $s_n$ and $s_n^{\prime}$ are the $n^{\text {th }}$ partial sums of the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$, respectively. By the assumption, we have $0 \leq s_n \leq s_n^{\prime}$ for all $n \in \mathbb{N}$, and both $\left(s_n\right)$ and $\left(s_n^{\prime}\right)$ are monotonically increasing.

Suppose $\sum_{n=1}^{\infty} b_n$ converges, that is, $\left(s_n^{\prime}\right)$ converges. Then, $\left(s_n^{\prime}\right)$ is bounded. Hence, by the relation $0 \leq s_n \leq s_n^{\prime}$ for all $n \in \mathbb{N},\left(s_n\right)$ is bounded as well as monotonically increasing. Therefore, by Theorem $1.1 .9,\left(s_n\right)$ converges.
The following corollary is immediate from the above theorem.

Corollary 1.2.7 (Comparison test) Let $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}$. Then
$$\sum_{n=1}^{\infty} a_n \text { diverges } \Rightarrow \sum_{n=1}^{\infty} b_n \text { diverges. }$$
Corollary 1.2.8 Suppose $\left(a_n\right)$ and $\left(b_n\right)$ are sequences of positive terms.
(i) Suppose $\ell:=\lim {n \rightarrow \infty} \frac{a_n}{b_n}$ exists. (a) If $\ell>0$, then $\sum{n=1}^{\infty} b_n$ converges $\Longleftrightarrow \sum_{n=1}^{\infty} a_n$ converges.
(b) If $\ell=0$, then $\sum_{n=1}^{\infty} b_n$ converges $\Rightarrow \sum_{n=1}^{\infty} a_n$ converges.
(ii) Suppose $\lim {n \rightarrow \infty} \frac{a_n}{b_n}=\infty$. Then $\sum{n=1}^{\infty} a_n$ converges $\Rightarrow \sum_{n=1}^{\infty} b_n$ converges.
Proof (i) Assume that $\ell:=\lim {n \rightarrow \infty} \frac{a_n}{b_n}$ exists, i.e., $0 \leq \ell<\infty$. (a) Suppose $\ell>0$. Then for any $0<\varepsilon<\ell$ there exists $n \in \mathbb{N}$ such that $$0 \leq \ell-\varepsilon<\frac{a_n}{b_n}<\ell+\varepsilon \quad \forall n \geq N .$$ Thus, $(\ell-\varepsilon) b_n{n=1}^{\infty} b_n$ converges if and only if $\sum_{n=1}^{\infty} a_n$ converges.
(b) Suppose $\ell=0$ and let $\varepsilon>0$ be given. Then, there exists $n \in \mathbb{N}$ such that $-\varepsilon<\frac{a_n}{b_n}<\varepsilon$ for all $n \geq N$. In particular,
$$a_n<\varepsilon b_n \quad \forall n \geq N .$$

## Calculus_微积分_Alternating Series

In the last subsection we have described some tests for asserting the convergence or divergence of series of non-negative terms. In this subsection we provide a sufficient condition for convergence of series with alternatively positive and negative terms.

Definition 1.2.3 A series of the form $\sum_{n=1}^{\infty}(-1)^{n+1} u_n$, where $\left(u_n\right)$ is a sequence of positive terms, is called an alternating series.
We have seen in Example 1.2.7 that the alternating series
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\cdots+\frac{(-1)^{n+1}}{n}+\cdots$$
is convergent. Note that the series
$$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{4 n-3}+\frac{1}{4 n-2}-\frac{1}{4 n-1}-\frac{1}{4 n}+\cdots$$
is not an alternating series. As you can see, in the latter case, though the series in not an alternating series, it is of the form
$$\sum_{n=1}^{\infty}\left[(-1)^{n+1} u_n+(-1)^{n+1} v_n\right]$$
with
$$u_n=\frac{1}{2 n-1}, \quad v_n=\frac{1}{2 n} .$$
We know from the results in Example 1.2.8 that the alternating series $\sum_{n=1}^{\infty}(-1)^{n+1} u_n$ and $\sum_{n=1}^{\infty}(-1)^{n+1} v_n$ are convergent. Hence, we can assert the convergence of the original series $\sum_{n=1}^{\infty}\left[(-1)^{n+1} u_n+(-1)^{n+1} v_n\right]$.

The next theorem, due to Leibnitz, ${ }^5$ provides such a sufficient condition for the convergence of alternating series.

## Calculus_微积分Some Tests for Convergence

$$\sum_{n=1}^{\infty} a_n \text { diverges } \Rightarrow \sum_{n=1}^{\infty} b_n \text { diverges. }$$

(我想 $\ell:=\lim n \rightarrow \infty \frac{a_n}{b_n}$ 存在。 $(\mathrm{a})$ 如果 $\ell>0$ ，然后 $\sum n=1^{\infty} b_n$ 收敛 $\Longleftrightarrow \sum_{n=1}^{\infty} a_n$ 收敛。
(b) 如果 $\ell=0$ ，然后 $\sum_{n=1}^{\infty} b_n$ 收敛 $\Rightarrow \sum_{n=1}^{\infty} a_n$ 收敛。
(ii) 假设 $\lim n \rightarrow \infty \frac{a_n}{b_n}=\infty$. 然后 $\sum n=1^{\infty} a_n$ 收敛 $\Rightarrow \sum_{n=1}^{\infty} b_n$ 收敛。

$$a_n<\varepsilon b_n \quad \forall n \geq N .$$

## Calculus_微积分Alternating Series

$$u_n=\frac{1}{2 n-1}, \quad v_n=\frac{1}{2 n} .$$

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