Let $x$ be a real number and $n$ a natural number greater than or equal to 1. Calculate:

$$
1+x+x^{2}+\cdots+x^{n}
$$
A case distinction must be made according to $x=1$ and $x \neq 1$ and we find:

$$
1+x+x^{2}+\cdots+x^{n}= \begin{cases}\frac{1-x^{n+1}}{1-x} & \text { if } x \neq 1 \\ n+1 & \text { if } x=1\end{cases}
$$

## - Inequalities -

## Basic Manipulations

## Theorem 1.1.5.

Let $a, b, c, d, \lambda$ be real numbers.

— Link between strict and non-strict: If $a<b$, then $a \leqslant b$. THE CONVERSE IS FALSE!

- Sum: If $\left\{\begin{array}{l}a \leqslant b \\ c \leqslant d\end{array}\right.$, then $a+c \leqslant b+d$.
- Product
- by a positive real: if $a \leqslant b$ and $\lambda \geqslant 0$, then: $\quad \lambda a \leqslant \lambda b$.
- by a negative real: if $a \leqslant b$ and $\lambda \leqslant 0$, then: $\lambda a \geqslant \lambda b$.

![](https://cdn.mathpix.com/cropped/2024_05_10_346641170ac76390eef3g-028.jpg?height=137&width=1234&top_left_y=2296&top_left_x=297)

- Square and square root:

$$
\begin{aligned}
0 \leqslant a \leqslant b & \Longleftrightarrow \sqrt{a} \leqslant \sqrt{b} \\
a^{2} \leqslant b^{2} & \Longleftrightarrow|a| \leqslant|b|
\end{aligned}
$$

## Triangular Inequalities

## Theorem 1.1.6.

Let $x$ and $y$ be two real numbers. Then we have the following inequalities:

$$
\begin{aligned}
|x+y| & \leqslant|x|+|y| \quad \text { (Triangular Inequality 1) } \\
|| x|-| y|| & \leqslant|x-y| \quad \text { (Triangular Inequality 2) }
\end{aligned}
$$

## Proof.

- Triangular Inequality 1: We compare the squares and the result follows, knowing that $x y \leqslant$ $|x y|$.
- Triangular Inequality 2: Triangular Inequality 2 amounts to showing that

$$
|x|-|y| \leqslant|x-y| \quad \text { and } \quad|y|-|x| \leqslant|x-y|
$$

We then have the following inequalities:

$$
|x|=|(x-y)+y| \underset{I T 1}{\leqslant}|x-y|+|y|
$$

## Technique: a square is positive!

& Explanation It is well known that a square is always positive:

$$
5^{2}=25 \text { is positive, }(-6)^{2}=36 \text { is positive, etc... }
$$

More generally, it is good to think of the following trivial theorem for establishing inequalities:

## Theorem 1.1.7.

A square is always positive, in other words:

$$
\text { for all real } x, \quad x^{2} \geqslant 0
$$