# Thread: Prove this is not an integer

1. ## Prove this is not an integer

Let $p_{i} \ \ \ \ \ 1 \leq i \leq n$ be from the first to the nth prime numbers, prove that $\frac {1}{p_{1}} + \frac {1}{p_{2}} + . . . + \frac {1}{p_{n}}$ is never an integer.

Proof.

I use induction, but I don't know if it is okay.

The first prime number is 2, 1/2 is not an integer, so 1 is okay.

Let $\sum ^{k}_{1} \frac {1}{p_{i}}$ be not an integer.

Then $\sum ^{k+1}_{1} \frac {1}{p_{i}} = \sum ^{k}_{1} \frac{1}{p_{i}} + \frac {1}{p_{k+1}}$ is also not an integer.

Let $p_{i} \ \ \ \ \ 1 \leq i \leq n$ be from the first to the nth prime numbers, prove that $\frac {1}{p_{1}} + \frac {1}{p_{2}} + . . . + \frac {1}{p_{n}}$ is never an integer.
Note,
$\frac{1}{p_1}+...+\frac{1}{p_n} = \frac{\sum_{k=1}^n \prod_{i=1,i\not =k}^n p_i}{\prod_{i=1}^n p_i}$.
If that is an integer then if we multiply it by $\prod_{i=1}^{n-1}p_i$ it will stay an integer.
If we do that we are left with,
$\frac{\sum_{k=1}^n \prod_{i=1,i\not =k}^n p_i}{p_n}$.
If we expand the summation there will be $n$ products each product will have a $p_n$ factor except the last one (when $k=n$). Thus, we have the first $n-1$ terms divisible by $p_n$ but not the last term, hence this cannot be an integer.