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?

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.  
