# Show that the sum $\frac {1}{p_1} + \frac {1}{p_2} +\frac {1}{p_3} +…+\frac {1}{p_n}$ is never an integer,where $p_i$'s are primes,$1\le i \le n$.

Assuming that $P_{n}$ is the n-th prime number, show that the sum $\frac {1}{p_1} + \frac {1}{p_2} +\frac {1}{p_3} +...+\frac {1}{p_n}$ is never an integer.

I have been struggle all day with this problem i do not even see how to begin. Any idea, helps please

We have, $P_1=2, P_2=3\dots$ and so on. I said this because, you should know that there is only one even prime i.e.$2$, all prime else are odd.

Now, consider the l.c.m. of $P_1, P_2,\dots ,P_n$, and call this $l$. Look at this, $l$ is an even number(it is simply $P_1\times P_2\times\dots=2\times 3\times\dots$) .

Now, consider the sum $$S=\frac l{P_1}+\frac l{P_2}+\dots+\frac l{P_n}.$$

Note that, $S$ is an odd number, because,in the summands, the only term $\frac l{P_1}$ is odd, as the factor $2$ of $l$ vanishes out, and others are even, because there the factor $2$ never vanishes.

Now, $$\frac Sl=\frac 1{P_1}+\frac 1{P_2}+\dots+\frac 1{P_n}.$$ Since $S$ is odd and $l$ is even, the fraction cannot be integer, so, your claim.

Hint: Multiply it by the product of those primes. Is the result a multiple of $p_1$?

To start, simply pick out constraints from the problem. For example, notice that there are finitely many primes, hence there is a largest one. Primes have a very simple prime factorization (and integer primes are indivisible). Further, a fraction that's an integer has a very special form.

Try doing some basic algebra to get a nicer looking equation -- an equation with some things being multiplied/added on one side and things being multiplied/added on the other side.