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 $s_{n+1}$.

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 $$s_{n+1}$$.

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?

• A proof of the proposition is claimed in the comments to OEIS007504 – Keith Backman Apr 10 at 18:19

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

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

• Is there any more elementary proof? This contains a lot of calculations... Besides, thanks for the information, I will look up the formula! – kelvin hong 方 Apr 10 at 9:11

After some discussion with my professor, we have solved this problem by using some elementary arguments.

We first prove the following lemma.

Lemma : For $$n\geq 4$$ we have $$s_n<\bigg(\frac{p_{n+1}-1}{2}\bigg)^2.$$ To show this we first see $$s_4=2+3+5+7=17$$ and $$(p_{5}-1)/2=5$$, so obviously $$17< 5^2$$. Now if there is $$m\geq 4$$ such that $$s_m<[(p_{m+1}-1)/2]^2$$, then $$s_{m+1}=s_m+p_{m+1}<\bigg(\frac{p_{m+1}-1}2\bigg)^2+p_{m+1}=\bigg(\frac{p_{m+1}+1}2\bigg)^2\leq \bigg(\frac{p_{m+2}-1}{2}\bigg)^2$$ since for primes $$\geq 11$$, every pair of consecutive primes has least gap equals to $$2$$, the lemma's proof is completed now.

Now choose $$n\geq 4$$, let $$a$$ be the largest integer satisfy $$a^2\leq s_n$$, then by definition we have $$(a+1)^2> s_n$$. Now we have $$a^2\leq s_n< \bigg(\frac{p_{n+1}-1}2\bigg)^2\implies a< \frac{p_{n+1}-1}{2}$$ therefore $$2a+1< p_{n+1}$$. Combining these we have $$(a+1)^2=a^2+2a+1< s_n+p_{n+1}=s_{n+1}$$ So we have proved that $$s_n< (a+1)^2< s_{n+1}$$.

We have $$s_1=2, s_2=5, s_3=10, s_4=17$$, so $$s_1< 2^2< s_2< 3^2< s_3< 4^2< s_4$$ completely proved this statement.