### 数学代写|微积分代写Calculus代写|Some Tests for Convergence

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## 数学代写|微积分代写Calculus代写|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, $(\ell-\varepsilon) b_{n}0$ 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,
$$a_{n}<\varepsilon b_{n} \quad \forall n \geq N .$$
Again, by comparison test (Theorem 1.2.6), convergence of $\sum_{n=1}^{\infty} b_{n}$ implies the convergence of $\sum_{n=1}^{\infty} a_{n}$.
(ii) Assume that $\frac{a_{s}}{b_{n}} \rightarrow \infty$. Then there exists $N \in \mathbb{N}$ such that $\frac{a_{s}}{b_{n}} \geq 1$ for all $n \geq N$, i.e.,
$$b_{n} \leq a_{n} \quad \forall n \geq N .$$
Hence, comparison test can be applied in this case as well to obtain the required result.

## 数学代写|微积分代写Calculus代写|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.

Theorem 1.2.13 (Leibnitz’s theorem) Suppose ( $\left.u_{n}\right)$ is a sequence of positive terms such that $u_{n} \geq u_{n+1}$ for all $n \in \mathbb{N}$ and $u_{n} \rightarrow 0$ as $n \rightarrow \infty$. Then the alternating series $\sum_{n=1}^{\infty}(-1)^{n+1} u_{n}$ converges, and in that case,
$$\left|s-s_{n}\right| \leq u_{n+1} \quad \forall n \in \mathbb{N}$$
where
$$s_{n}=\sum_{j=1}^{n}(-1)^{j+1} u_{j} \quad \text { and } \quad s=\sum_{n=1}^{\infty}(-1)^{n+1} u_{n}$$
Proof We observe that
$$s_{2 n+1}=s_{2 n}+u_{2 n+1} \quad \forall n \in \mathbb{N}$$

The convergence of the series
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+\frac{(-1)^{n}}{2 n+1}+\cdots$$
which is generally known as Leibnitz-Gregory series, was known to Indian mathematicians as early as in 15 -th century, and the value of the above series was proved to be $\frac{\pi}{4}$. The above series appeared in the work of a Kerala mathematician Madhava around 1425 that was presented later in the year around 1550 by another Kerala mathematician Nilakantha (cf. [7]). The discovery of the above series is normally attributed to Leibnitz and James Gregory after nearly 300 years of its discovery. Respecting the chronology of its discovery, we shall refer this series as MadhavaNilakantha series.

Let us give a simple proof for the equality
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+\frac{(-1)^{n+1}}{2 n-1}+\cdots$$
using some elementary rules of integration that one studies in school, which we shall study in detail in Chap. 6 .
We know that $(1-r)\left(1+r++\cdots+r^{n}\right)=\left(1-r^{n+1}\right)$ so that for $r \neq 1$,
$$\frac{1}{1-r}=1+r+\cdots+r^{n}+\frac{r^{n+1}}{1-r}$$
Now, taking $r=-x^{2}$ we have
$$\frac{1}{1+x^{2}}=1-x^{2}+x^{4}-x^{6}+\cdots+(-1)^{n} x^{2 n}+(-1)^{n+1} \frac{x^{2 n+2}}{1+x^{2}} .$$
On integration
$$\int \frac{d x}{1+x^{2}}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots+(-1)^{n} \frac{x^{2 n+1}}{2 n+1}+\int \frac{(-1)^{n+1} x^{2 n+2}}{1+x^{2}} d x$$
Now, recalling
$$\int_{0}^{1} \frac{d x}{1+x^{2}}=\tan ^{-1}(1)=\frac{\pi}{4},$$
we have
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+(-1)^{n} \frac{1}{2 n+1}+\int_{0}^{1} \frac{(-1)^{n+1} x^{2 n+1}}{1+x^{2}} d x$$
Now, observe that
$$\left|\int_{0}^{1}(-1)^{n+1} \frac{x^{2 n+2}}{1+x^{2}} d x\right| \leq \int_{0}^{1} \frac{x^{2 n+2}}{1+x^{2}} d x \leq \int_{0}^{1} x^{2 n+2} d x=\frac{1}{2 n+3}$$
Thus,
$$\left|\frac{\pi}{4}-\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+(-1)^{n} \frac{1}{2 n+1}\right)\right| \leq \frac{1}{2 n+3} \rightarrow 0$$
Thus, we have proved that $$\frac{\pi}{4}=1+\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2 n+1}$$
Using the procedure used above and the fact that $\int_{0}^{1} \frac{d x}{1+x}=\log 2$, we see that
$$\log 2=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$$
Exercise 1.2.6 Derive the above series representation for $\log 2$.

## 数学代写|微积分代写Calculus代写|Some Tests for Convergence

∑n=1∞bn 收敛 ⇒∑n=1∞一个n 收敛

∑n=1∞一个n 分歧 ⇒∑n=1∞bn 分歧。

（我想ℓ:=林n→∞一个nbn存在。
(a) 如果ℓ>0， 然后∑n=1∞bn收敛⟺∑n=1∞一个n收敛。
(b) 如果ℓ=0， 然后∑n=1∞bn收敛⇒∑n=1∞一个n收敛。
(ii) 假设林n→∞一个nbn=∞. 然后∑n=1∞一个n收敛⇒∑n=1∞bn收敛。

0≤ℓ−e<一个nbn<ℓ+e∀n≥ñ.因此，(ℓ−e)bn0被给予。那么，存在n∈ñ这样−e<一个nbn<e对所有人n≥ñ. 尤其是，

(ii) 假设一个sbn→∞. 那么存在ñ∈ñ这样一个sbn≥1对所有人n≥ñ， IE，

bn≤一个n∀n≥ñ.

## 数学代写|微积分代写Calculus代写|Alternating Series

1−12+13−14+15+⋯+(−1)n+1n+⋯

1+12−13−14+⋯+14n−3+14n−2−14n−1−14n+⋯

∑n=1∞[(−1)n+1在n+(−1)n+1在n]

|s−sn|≤在n+1∀n∈ñ

sn=∑j=1n(−1)j+1在j 和 s=∑n=1∞(−1)n+1在n

s2n+1=s2n+在2n+1∀n∈ñ

1−13+15−17+⋯+(−1)n2n+1+⋯

11−r=1+r+⋯+rn+rn+11−r

11+X2=1−X2+X4−X6+⋯+(−1)nX2n+(−1)n+1X2n+21+X2.

∫dX1+X2=X−X33+X55−X77+⋯+(−1)nX2n+12n+1+∫(−1)n+1X2n+21+X2dX

∫01dX1+X2=棕褐色−1⁡(1)=圆周率4,

|∫01(−1)n+1X2n+21+X2dX|≤∫01X2n+21+X2dX≤∫01X2n+2dX=12n+3

|圆周率4−(1−13+15−17+⋯+(−1)n12n+1)|≤12n+3→0

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