But once in a while we meet a property which is amenable to a simple proof. Here is one such property.
Let $p_n$ denote the $n$-th prime number. The sequence of primes is
\[\left\langle p_n \right \rangle_{n=1}^{\infty} \ = \ \left \langle 2, \, 3, \, 5, \, 7, \, 11, \, 13, \, 17, \, 19, \, 23, \, 29, \, 31, \, \ldots\right\rangle.\]
Let us form the partial sums of this sequence, i.e., the sums
\[2, \, 2+3, \, 2+3+5, \, 2+3+5+7, \, 2+3+5+7+11, \, \ldots.\]
We get the following sequence of numbers, which we denote by $\left \langle a_n \right \rangle_{n=1}^{\infty}$, so that $a_n = p_1 +p_2 + \cdots + p_n = \sum_{i=1}^n p_i$:
\[\left\langle a_n \right\rangle_{n=1}^{\infty} \ = \ \left\langle 2, \, 5, \, 10, \, 17, \, 28, \, 41, \, 58, \, 77, \, 100, \, 129, \, 160, \, \ldots\right\rangle.\]
Here is an interesting property of the above sequence of partial sums:
- Between every two numbers in the sequence there exists a perfect square number.
In other words, for each positive integer $n$, there exists an integer $k$ such that
\[a_n < k^2 < a_{n+1}.\]
Note that we are not asserting that there is just one such number: there may well be more than one square number between $a_n$ and $a_{n+1}$. Indeed, a counterexample to such a claim (were anyone to make it) is easily spotted: we find that $a_{11} = 160$ and $a_{12} = 197$, and between the two numbers $160$ and $197$ there lie two square numbers, $169$ and $196$.
Proof, anyone?
