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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
$$\int \frac{d x}{1+x^2}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots+(-1)^n \frac{x^{2 n+1}}{2 n+1}+\int \frac{(-1)^{n+1} x^{2 n+2}}{1+x^2} d x$$

## Calculus_微积分_Absolute Convergence

We know that for a sequence $\left(a_n\right)$, the series $\sum_{n=1}^{\infty} a_n$ may converge, but $\sum_{n=1}^{\infty}\left|a_n\right|$ can diverge. For example we have seen that $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges whereas $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.

Definition 1.2.4 Let $\left(a_n\right)$ be a sequence of real numbers. Then the series $\sum_{n=1}^{\infty} a_n$ is said to be
(1) absolutely convergent, if $\sum_{n=1}^{\infty}\left|a_n\right|$ is convergent,
(2) conditionally convergent, if it converges, but not absolutely.
Example 1.2.18 (i) the series
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}, \quad \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !}, \quad \sum_{n=1}^{\infty} \frac{\sin n}{n^2}$$
are absolutely convergent, so also the series
$$\sum_{n=1}^{\infty} \frac{a^n}{n !}$$
for any $a \in \mathbb{R}$. (ii) We already observed in Sect. 1.2.4 that the series
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \text { and } \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n-1}$$
are convergent series, which also from Leibnitz theorem, but they are not absolutely convergent. Thus, these series are conditionally convergent.

## Calculus微积分Absolute Convergence

$\sum{n=1}^{\infty} \frac{1}{n}$ 分歧。

(1) 缁对收敛, 如果 $\sum_{n=1}^{\infty}\left|a_n\right|$ 是收敛的，
（2）条件收玫，如果收敛，但不是绝对收玫。

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}, \quad \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !}, \quad \sum_{n=1}^{\infty} \frac{\sin n}{n^2}$$

$$\sum_{n=1}^{\infty} \frac{a^n}{n !}$$

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \text { and } \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n-1}$$

## MATLAB代写

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