Is my proof mathematically correct?

Question

I could use some expert advice/ suggestions here to formulate the proof properly. Evaluate:
$$\lim _{N\rightarrow \infty } \left( \,{\frac {{N}^{3}}{8\sum _{k=0}^ {N-1}({k}^{2}+\left\lfloor k/2\right\rfloor)}} \right) $$
First we inspect the numerator and denominator. The numerator itself is a continuous function albeit the range for input values: N∈ ℕ+. The denominator contains the operator $\lfloor$ $\rfloor$ (floor) and should be eliminated.
We can rewrite the denominator within the summation as follows:

$${k}^{2}+\left\lfloor k/2\right\rfloor={k}^{2}+k/2+O \left( r \right) $$

, where O(r) is a remainder.
We now define the error ε as: $\varepsilon={\frac {O \left( r \right) }{{k}^{2}+\left\lfloor k/2\right\rfloor}}$
Then the limit for ε as k goes to infinity is:

$$\lim _{k\rightarrow \infty }{\frac {O \left( r \right) }{{k}^{2}+\left\lfloor k/2\right\rfloor}}=0$$
We have established that the operator $\lfloor$ $\rfloor$, for convergence analysis, can be eliminated. In the denominator, we now have a continuous function:

$$\lim _{N\rightarrow \infty } \left( {\frac {{N}^{3}}{8\,\sum _{k=0}^{N-1}({k}^{2}+k/2)}} \right) =\lim _{N\rightarrow\infty }{\frac {{N}^{3}}{8/3\,{N}^{3}-2\,{N}^{2}-2/3\,N}}$$
Then we divide with ${N}^{3}$ in both the numerator and denominator.

$$\lim _{N\rightarrow \infty }{\frac {{1}}{8/3\,-2\,{N}^{-1}-(2/3)\,{N}^{-2}}}=3/8$$
End of proof.

Answer 1

Your result is correct, though perhaps the use of remainder here (what I called "fractional part" in the following) is a little unusual in my opinion, in particular because you denoted that as $\;O(r)\;$ and it is unclear what $\;r\;$ is, how it depends on $\;N\;$ and why the limit of $\;\epsilon\;$ is zero if we don't know those things.

I would propose the following alternative way using the fractional part function $\;\{x\}:=x-\lfloor x\rfloor\;$:

$$k^2+\left\lfloor\frac k2\right\rfloor=k^2+\frac k2-\left\{\frac k2\right\}$$

and then

$$\sum_{k=0}^{N-1}\left(k^2+\left\lfloor\frac k2\right\rfloor\right)=\sum_{k=0}^{N-1}\left(k^2+\frac k2-\left\{\frac k2\right\}\right)=$$

$$=\underbrace{\frac{(N-1)N(\overbrace{2(N-1)+1}^{=2N-1})}6+\frac{(N-1)N}4}_{=\cfrac{N(N-1)(4N+1)}{12}}-\sum_{k=0}^{N-1}\left\{\frac k2\right\}$$

But

$$\left\{\frac k2\right\}=\begin{cases}0,&k=2m\\{}\\\cfrac12,&m=2k+1\end{cases}\;,\;\;m\in\Bbb Z$$

so the last sum above is at most $\;\cfrac12\cdot N=\cfrac N2\;$ , and thus:

$$\frac{N(N-1)(4N+1)}{12}-\frac N2\le\sum_{k=0}^{N-1}\left(k^2+\left\lfloor\frac k2\right\rfloor\right)\le\frac{N(N-1)(4N+1)}{12}$$

and from here:

$$8\frac{\frac{N(N-1)(4N+1)}{12}-\frac N2}{N^3}\le\frac{8\sum\limits_{k=0}^{N-1}\left(k^2+\left\lfloor\frac k2\right\rfloor\right)}{N^3}\le8\frac{\frac{N(N-1)(4N+1)}{12}}{N^3}$$

and now it is easy to show each extreme's limit is $\;\cfrac83\;$ , so your sequence's limit indeed is $\;\cfrac38\;$

Answer 2

In the summation, $k^2$ yields an $O(N^3)$ contribution, while $\lfloor k/2\rfloor$, which is bounded by $k/2$, only yields an $O(N^2)$ contribution and can be neglected.

The sum can be computed as a Riemannian integral,

$$\lim_{N\to\infty}\frac1N\sum_0^{N-1}\left(\frac kN\right)^2=\int_0^1x^2\,dx.$$ Hence the answer $$\frac38.$$

Answer 3

One has $$\sum_{k=0}^{n-1}k^2={(n-1)n(2n-1)\over 6}={n^3\over3}\bigl(1+o(1)\bigr)\qquad(n\to\infty)$$ and $$0\leq\sum_{k=0}^{n-1}\lfloor k/2\rfloor\leq{(n-1)n\over4}={n^3\over3}\>o(1)\qquad(n\to\infty)\ .$$ It follows that $${8\sum_{k=0}^{n-1}\bigl(k^2+\lfloor k/2\rfloor\bigr)\over n^3}={8n^3\bigl(1+o(1)\bigr)\over 3\>n^3}\to{8\over3}\qquad(n\to\infty)\ ,$$ hence your limit is ${3\over8}$.