Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$.

Question

Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$.

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I cannot give a proof to this, although I have try on some small examples.

I also notice that $\pi(x)\sim x/\log x$ and there are approximately $\sqrt x$ perfect square smaller than $x$. I have a feel that because $\sqrt x=\mathcal O(x/\log x)$, there will be more perfect squares, two or three, between $s_n$ and $s_{n+1}$ when $n$ gets large.

Any suggestion?

Answer 1

The sum of the first $n$ primes is

$$ s_n = \frac{n^2}{2}\Big(\log n + \log\log n - \frac32 + \frac{\log\log n}{\log n} - \frac{5}{2\log n} + o(1)\Big) $$

Which gives $$ s_{n+1} - s_n \approx n^2 (\log n + \log\log n - 3/2) > n^2 $$

for all $n > 10$. For $n = 1,2,3$ we can verify it manually. Since the gap between $s_n$ and $s_{n+1}$ jumps by more than $n^2$ the interval $s_n, s_{n+1}$ will contains a square. More specifically we can prove that

$$ s_n < (n+1)^2 < s_{n+1} $$