TITLE: Calculus (Limits) Doubt: $\theta - \cfrac{\theta^3}{3!} < \sin \theta < \theta$ use to solve limit.
QUESTION [5 upvotes]: Following is the question I've been trying to work on but can't get enough of it: 

$$\lim_{n\rightarrow \infty} \sin\left(\cfrac{n}{n^2+1^2}\right) + \sin\left(\cfrac{n}{n^2+2^2}\right) + \cdots + \sin\left({\cfrac{n}{n^2+n^2}}\right) $$

I'm required to find the value of the above limit. All I could think about is to take $n^2$ common from numerator and denominator of each term. 
$$\lim_{n\rightarrow \infty} \sin\left(\cfrac{1/n}{1+1^2/n^2}\right) + \sin\left(\cfrac{1/n}{1+2^2/n^2}\right) + \cdots + \sin\left({\cfrac{1/n}{1+n^2/n^2}}\right) $$
Now since $n \to \infty$ then shouldn't each term inside the sine function be zero and thus value of limit be zero? Where am I going wrong in this approach? 
Also, I found a trick for this question specified in my book as:
To use the following inequality : $$\color{blue}{\theta - \cfrac{\theta^3}{3!} < \sin \theta < \theta }$$
And then to replace $\theta$ with $\cfrac{n}{n^2+k^2}$. I've never seen this inequality before, can anyone refer to the proof of this inequality? (or give the proof). 
EDIT: 
After working with the suggestions posted in the comments and answers: (and the link of the duplicate post)
I do get the following equation: 
$$ \lim_{n \to \infty} \sum_{k=1}^{n} \cfrac{1}{n} \left(\cfrac{1}{1+(k/n)^2}\right)$$
How can I convert this into Integral now?

REPLY [3 votes]: (Proof without Taylor series)
First we prove the right side.
$$\begin{align}
&f(x)=\sin x, \quad g(x)=x\\
&f(0)=g(0)\\
&f'(x)=\cos x \leq 1=g'(x)\\
&\therefore f(x)\leq g(x)
\end{align}$$
Now we prove the left side.
$$\begin{align}
&h(x)=x-\frac{x^3}{3!},\quad f(x)=\sin x\\
&h(0)=f(0)\\
&f'(x)-h'(x)=\cos x-\left(1-\frac{x^2}{2}\right)\\
&f'(0)-h'(0)=0\\
&f''(x)-h''(x)=-\sin x+x\geq 0 \text{ (we proved it above)}\\
&\therefore h'(x)\leq f'(x)\quad\text{and}\quad h(x)\leq f(x)
\end{align}$$