If $L=\lim_{m\to \infty}\sum_{p=1}^m \frac{p}{2p+m+m^2}$, then find floor(L). 
If $L=\displaystyle\lim_{m\to \infty}\sum_{p=1}^m \frac{p}{2p+m+m^2}$, then find floor(L).

This is from a preparatory examination for the college entrance exams for high-school students. 
I first tried to convert it into a definite integral by the conventional way, but the $m^2$ term kept interfering. Now the only other thing that I could think of was the sandwich theorem or the squeeze theorem, for which also I could think about only the upper bound. This was $$L\leq \lim_{m\to \infty}\sum_{p=1}^m \frac{p}{2p+m}= \int_0^1\frac{xdx}{2x+1}=\frac{2-\ln 3}{4}$$ while this gives me the correct value of the floor function, but the solution claims that $L=\frac 12$.
 A: Note that:
$$ \sum_{p=1}^{m} \frac{p}{{\color{Red}{2m}}+m+m^2} \leq \sum_{p=1}^{m} \frac{p}{{\color{Red}{2p}}+m+m^2} \leq \sum_{p=1}^{m} \frac{p}{{\color{Red}{1}}+m+m^2}$$
$$\frac{m(m+1)}{2(2m+m+m^2)} \leq \sum_{p=1}^{m} \frac{p}{2p+m+m^2} \leq \frac{m(m+1)}{2(1+m+m^2)} $$
As $m \to \infty$, the sum tends to $\dfrac 12$.
A: Here is an idea.
$$
\sum_{p=1}^m \frac{p}{2p+m+m^2}=\sum_p\frac{p/m}{2(p/m)+1+m}=\sum_p\frac{p/m}{m(\frac{2(p/m)+1}{m}+1)}=\\\sum_p\sum_{k=0}^\infty\frac{1}{m}(-1)^k p/m(2p/m+1)^k\times \frac{1}{m^k}
$$
I don't know how rigorous you have to be but you can exchange both sums and take the limit term by term. then without the $\frac{1}{m^k}$ the $p$-sum converges to an integral.
$$
\int_0^1dt \, t(2t+1)^k<3^k
$$
Hence the $k$-th term goes as $3^k/m^k\to 0$ for $k>0$ so you only have to take care of the term of the $k=0$ term. This one gives $\int_0^1 t=1/2$.
The problem with your solution is that
$$
\lim_{m\to \infty}\sum_{p=1}^m\frac{p}{2p+m}\neq \lim_{m\to \infty}\frac{1}{m}\sum_{p=1}^m\frac{p/m}{2p/m+1}=\int_0^1\frac{t}{2t+1}dt
$$
