# Finding the upper and lower bound before squeeze theorem

I want to find the limit of the following sequence. I know that the second part of this question involves using the squeeze theorem yet I have no idea how to go about getting the upper and lower bounds. I've looked at other questions and am generally very confused so if you can make any help as simple as possible that would be great.

$$a_n = (n^2 + 1)\sin\Bigl(\,\frac{1}{n^3+1}\Bigr) + (n^2 + 2)\sin\Bigl(\,\frac{1}{n^3+2}\Bigr) + \dots +(n^2 + n)\sin\Bigl(\,\frac{1}{n^3+n}\Bigr)$$

Using the following inequality : $x-\frac{x^3}{6}\le sin(x) \le x \ \forall x\ge 0$ we get : $\sum_{k=1}^{n}(n^2+k)(\frac{1}{n^3+k}-\frac{1}{6(n^3+k)^3})\le a_n\le \sum_{k=1}^n\frac{n^2+k}{n^3+k}$ and then using the fact that $1\le k \le n$ we get : $\sum_{k=1}^{n}(n^2+k)(\frac{1}{n^3+n}-\frac{1}{6(n^3+1)^3})\le a_n\le \sum_{k=1}^n\frac{n^2+k}{n^3+1}$ and so
$\frac{n^3}{n^3+n}-\frac{n^3}{6(n^3+1)^3}+\frac{n+1}{2(n^2+1)}-\frac{n(n+1)}{12(n^3+1)^3}\le a_n \le\frac{n^3}{n^3+1}+\frac{n(n+1)}{2(n^3+1)}$
Both sequences converge to $1$ and so $a_n$ converges to $1$ as well.