1. ## Sum of reciprocals

Hii !

Can You help me, for this Exercice :

Prove that for all integer $\displaystyle p \geq 3$ it exist p of integers natural different two to two $\displaystyle n_1 , n_2 , ... , n_p$ Such as :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$

2. Originally Posted by Perelman
Hii !

Can You help me, for this Exercice :

Prove that for all integer $\displaystyle p \geq 3$ it exist p of integers natural different two to two $\displaystyle n_1 , n_2 , ... , n_p$ Such as :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$
Come again?

3. Wath VonNemo19 ?

thanks in advance for any help ...

4. Originally Posted by Perelman
Hii !

Can You help me, for this Exercice :

Prove that for all integer $\displaystyle p \geq 3$ it exist p of integers natural different two to two $\displaystyle n_1 , n_2 , ... , n_p$ Such as :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$
What you have posted makes no sense. Please post the question exactly as it's written from where you got it from.

5. Are you French ???

6. Originally Posted by Moo
Are you French ???
Exactly . by this I have some problemes to traduct ..

7. Oui, j'ai cru voir ça. Pose ta question en Français, je vais (essayer de) traduire
Sinon, il existe des forums de maths en Français lol, ce qui me paraît être la meilleure solution. Et c'est quoi ton niveau ? Parce que ce ne sont pas des probabilités ça !

8. Originally Posted by Perelman
Hii !

Can You help me, for this Exercice :

Prove that for all integer $\displaystyle p \geq 3$ it exist p of integers natural different two to two $\displaystyle n_1 , n_2 , ... , n_p$ Such as :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$
This is called "Egyptian fractions" ("fractions égyptiennes"); tu devrais taper ça dans Google, il y a de nombreux sites qui en parlent.

9. Voiçi, la question en Français :

prouver que pour tout entier $\displaystyle p \geq 3$ il existent p d'entiers naturels différents deux à deux $\displaystyle n_1 , n_2 , ... , n_p$ tels que :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$

10. The question is (for those who're interested ..) :

Prove that for any integer p $\displaystyle \geq 3$, there exist p pairwise different positive integers $\displaystyle n_1,n_2,\dots,n_p$ such that :

$\displaystyle \frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_p} = 1$

Look at what Laurent referred you to, it should be useful to solve your problem.