# How to show $p_n < p_1 + p_2 + \cdots + p_{n-1}$?

How do you show that, if $p_n$ denotes the $n$th prime, $n > 3$, then

$$p_n < p_1 + p_2 +\cdots + p_{n-1}$$

using the Bertrand conjecture and induction?

Here is what I've come up with, but I'm not sure if it's correct:

Let $n=4$. Then, $p_4 = 7$, so we have:

$$7 < 2+3+5 = 10.$$

Assume that the statement is true for $n$ so that $p_n < p_1 + p_2 +\cdots + p_{n-1}$.

We want to prove that this is true for $n+1$.

$$p_{n+1} < p_1 + p_2 +\dotsc + p_{n-1} + p_n.$$

But we assumed that $p_n < p_1 + p_2 +\cdots + p_{n-1}$, so we have:

$$p_{n+1} < p_1 + p_2 +\cdots + p_{n-1} + p_n < p_n + p_n = 2p_n.$$

We know that $p_n < p_{n+1}$, so we have:

$p_n < p_{n+1} < 2p_n$, which is true by the Bertrand conjecture. $\blacksquare$

• You have a good idea about relating this problem to Bertrand's "postulate", but you get some steps out of order toward the end of your argument. Go through the next to last line a little slower, and keep in mind that you know $p_{n+1} \lt 2 p_n$, and are trying to prove $p_{n+1} \lt p_1 + \ldots +p_n$. – hardmath Sep 25 '15 at 18:41

$$p_{n+1} < p_1 + p_2 +\dotsc + p_{n-1} + p_n < p_n + p_n = 2p_n$$
the second inequality is wrong. It would be better to first invoke Bertrand's postulate to establish $p_{n+1}\lt 2p_n$ and then apply the inductive hypothesis to get
$$p_{n+1}\lt2p_n=p_n+p_n\lt p_1+p_2+\cdots+p_{n-1}+p_n$$