<s>[INST] $\lim_{n\to\infty} \dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$ using the Squeeze Theorem Evaluate 

$$\lim_{n\to\infty} \dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$$

I know that the standard way would be to use the Closed Form of$$1^3+2^3+3^3+...+n^3$$ However I've been trying to practise using the Squeeze Theorem and thus tried to solve it using the Squeeze Theorem. However, I seem to have made a mistake as I don't get the expected answer.
$$$$My approach:$$$$
Taking $$f(n)=\dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$$ I identified the general term of $f(n)$ as $\dfrac{k^3}{(n^2+1)^2}$ where $1\le k\le n$
$$$$
Clearly $$\dfrac{1^3}{(n^2+1)^2}\le \dfrac{k^3}{(n^2+1)^2}\le \dfrac{n^3}{(n^2+1)^2}$$
Thus $$\dfrac{1^3}{(n^2+1)^2}+\dfrac{1^3}{(n^2+1)^2}+... \text{n times}\le f(n)\le \dfrac{n^3}{(n^2+1)^2}+\dfrac{n^3}{(n^2+1)^2}+... \text{n times}$$
$$\dfrac{n}{(n^2+1)^2}\le f(n)\le \dfrac{n\times n^3}{(n^2+1)^2}$$
$$\dfrac{n}{(n^2+1)^2}\le f(n)\le \dfrac{n^4}{(n^2+1)^2}$$
$$$$However on taking the limits as $n\to \infty$, I'm not getting the correct answer.
$$$$Could somebody please show me where I've gone wrong? Thanks!
 [/INST] Hint: 
$$\int_{0}^nx^3 dx<\sum_{i=1}^ni^3<n^3+\int_0^nx^3dx$$
$$\frac{1}{4}n^4<\sum_{i=1}^ni^3<n^3+\frac{1}{4}n^4$$
$$\frac{1}{4}\frac{n^4}{(n^2+1)^2}<\frac{\sum_{i=1}^ni^3}{(n^2+1)^2}<\frac{1}{4}\frac{4n^3+n^4}{(n^2+1)^2}$$
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