Proving an inequality of three sequences Let two sequences ($u_n$) and ($v_n$):
\begin{align}
u_n &= \sin(1/n^2) + \sin(2/n^2) + \ldots + \sin(n/n^2)\\[0.2cm]
v_n &= 1/n^2 + 2/n^2 + .... + n/n^2
\end{align}
In the previous parts of my exercise, I proved that $v_n$ is decreasing, convergent and its limit towards $+\infty$ is $\frac12$. I also proved that the following functions are all increasing and strictly positive in the interval $(0,+\infty)$
\begin{align}
f(x) &= x - \sin (x)\\
g(x) &= -1 + \tfrac12x^2 + \cos (x)\\
h(x) &= -x + \tfrac16x^3 + \sin (x)
\end{align}
Then, using these informations, I must prove the following inequality:
$$v_n - \frac16 \cdot\frac1{n^2} \leq u_n \leq v_n$$
before inferring that $u_n$ is convergent and $u_n$'s limit.
The problem is that I really don't know how to manage this part of my exercise : how to prove this inequality of sequences?
Thanks for your answers.
 A: You have done the difficult part which is to calculate the limit of the $v_n$ and to show that $h(x),f(x)\ge0$. Just as an observation: $h'(x)=g(x)$ and $h''(x)=g'(x)=f(x)$. Now, for any $1\le k\le n$ $$0\le h(k/n^2)=-\frac{k}{n^2}+\frac16\cdot\frac{k^3}{n^6}+\sin{\left(k/n^2\right)} \implies \frac{k}{n^2}-\frac{k^3}{6n^6}\le\sin{\left(k/n^2\right)}$$ and summing up over $k$ \begin{align}\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^3}{6n^6}\right)\le \sum_{k=1}^n\sin{\left(k/n^2\right)}&\implies v_n-\frac{1}{6n^6}\cdot\frac{n^2(n+1)^2}{4}\le u_n \\[0.2cm]&\implies v_n-\frac16\cdot\frac{1}{n^2}\le u_n\end{align} Similarly $$0\le f\left(k/n^2\right)=k/n^2-\sin{\left(k/n^2\right)}\implies \sum_{k=1}^{n}\sin(k/n^2)\le \sum_{k=1}^nk/n^2 \implies u_n\le v_n$$ Putting these together and using the Squeeze Theorem you have that \begin{align}\lim_{n\to +\infty }\left(v_n-\frac16\cdot\frac1{n^2}\right)\le \lim_{n\to+\infty}u_n\le \lim_{n\to+\infty}v_n&\implies \frac12-0\le \lim_{n\to+\infty}u_n\le \frac12 \\[0.2cm]&\implies\lim_{n\to+\infty}u_n=\frac12\end{align}

Although the result already obtains with the lower bound $$v_n-\frac{1}{6n^6}\frac{n^2(n+1)^2}{4}$$ it is only a straightforward calculation to show that $$v_n-\frac{1}{6n^2}\le v_n-\frac{1}{6n^6}\frac{n^2(n+1)^2}{4}$$ for any $n\ge 1$ in order to bring it in the form that the exercise requires.
