Calculus (Limits) Doubt: $\theta - \cfrac{\theta^3}{3!} < \sin \theta < \theta$ use to solve limit.

Question

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?

Answer 1

The problem with saying that "each term goes to 0, so the sum goes to 0" is that while each individual term goes to 0, the number of terms goes to infinity, so you wind up with (approximately) a limit of $0 \times \infty$, meaning you have to do something more to get the limit to work.

As to the approximation, it comes from the Taylor series expansion of $\sin{\theta} = \theta - \frac{1}{3!} \theta^3 + \frac{1}{5!} \theta^5 - \dots$, where you can get the two inequalities by truncating the series and looking at the error term - although it will only be the case for positive $\theta$.

Answer 2

(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. $$\begin{align} &h(x)=x-\frac{x^3}{3!},\quad f(x)=\sin x\\ &h(0)=f(0)\\ &f'(x)-h'(x)=\cos x-\left(1-\frac{x^2}{2}\right)\\ &f'(0)-h'(0)=0\\ &f''(x)-h''(x)=-\sin x+x\geq 0 \text{ (we proved it above)}\\ &\therefore h'(x)\leq f'(x)\quad\text{and}\quad h(x)\leq f(x) \end{align}$$

Answer 3

Hint

As pointed out in comments, the Taylor Expansion for $\sin(x)$ is $$\sin(\theta) = \theta - \frac{\theta^3}{3!} + O(\theta^5) > \theta - \frac{\theta^3}{3!}$$