$\lim_{n\to\infty} \dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$ 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}$$
$$$$However on taking the limits as $n\to \infty$, I'm not getting the correct answer.
$$$$Could somebody please show me where I've gone wrong? Thanks!
 A: We have $$(r + 1)^{4} - r^{4} = 4r^{3} + 6r^{2} + 6r + 1 > 4r^{3}\tag{1}$$ and $$r^{4} - (r - 1)^{4} = 4r^{3} - 6r^{2} + 6r - 1 < 4r^{3}\tag{2}$$ and putting $r = 1, 2, \ldots, n$ and adding these equations we get $$\frac{n^{4}}{4} < S = \sum_{r = 1}^{n}r^{3} < \frac{(n + 1)^{4} - 1}{4}\tag{3}$$ and then we can see that squeeze theorem can be easily applied to get desired limit as $1/4$. 

Update: OP is aware of the exact formula for sum $$S = \sum_{r = 1}^{n}r^{3}$$ and hence he knows that the limit is $1/4$. Since we are supposed to use the Squeeze theorem we must be able to bound the sum $S$ with two functions $f(n), g(n)$ such that $$f(n) < S < g(n)$$ and $$\frac{f(n)}{(n^{2} + 1)^{2}} \to \frac{1}{4}$$ and $$\frac{g(n)}{(n^{2} + 1)^{2}} \to \frac{1}{4}$$ Thus we want the bounds $f(n), g(n)$ to be of type $n^{4}/4$. In your approach the bounds are either too high or too low. Thus the upper bound $g(n)$ for $S$ in your answer looks like $n^{4}$ and lower bound $f(n)$ for $S$ looks like $n$. So from this you can only conclude that desired limit lies between $0$ and $1$.
Now the procedure to get the bounds comes from the technique used to obtain the closed form for $S$. This is done by calculating the differences $$(r + 1)^{4} - r^{4}$$ and $$r^{4} - (r - 1)^{4}$$ Any of the relations $(1)$ or $(2)$ of my answer can be used to derive an exact closed form for $S$, but we only need an approximation so the identities in $(1), (2)$ are given the shape of inequalities by neglecting terms containing $r^{2}, r$ and constant term. By summing these inequalities for $r = 1, 2, \ldots, n$ I don't get an exact formula for $S$ but rather very close bounds for $S$ and that's what we need here.
A: It is an overkill, but the Hermite-Hadamard inequality instantly solves the problem.
$f(x)=x^3$ is a convex function on $\mathbb{R}^+$, hence:
$$\frac{1}{4}=\int_{0}^{1}x^3\,dx \leq \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^3 \tag{1} $$
as well as:
$$ \frac{1}{n}\sum_{k=1}^{n+1}\left(\frac{k}{n}\right)^3 \leq \frac{1}{1+\frac{1}{n}}\int_{0}^{1+\frac{1}{n}}x^3\,dx = \frac{1}{4}\left(1+\frac{1}{n}\right)^3\tag{2} $$
where the difference between the LHS of $(2)$ and the RHS of $(1)$ is $O\left(\frac{1}{n}\right)$, so squeezing gives that the wanted limit is indeed $\color{red}{\large\frac{1}{4}}$, since $\lim_{n\to +\infty}\frac{n^4}{(n^2+1)^2}=1$.

An alternative approach is given by the combinatorial identity:
$$ \sum_{k=0}^{N}\binom{M+k}{M}=\binom{M+N+1}{M+1}. \tag{3}$$
For instance,
$$ 6\binom{n+1}{3}\leq n^3 \leq 6\binom{n+2}{3},\tag{4} $$
hence:
$$ 6\binom{N+2}{4} \leq \sum_{n=1}^{N} n^3 \leq 6\binom{N+3}{4}\tag{5} $$
If we multiply both sides of $(5)$ by $\frac{1}{(N^2+1)^2}$ and take the limit as $N\to +\infty$ we recover the previous result, always by squeezing.
A: Hint: 
$$\int_{0}^nx^3 dx<\sum_{i=1}^ni^3<n^3+\int_0^nx^3dx$$
$$\frac{1}{4}n^4<\sum_{i=1}^ni^3<n^3+\frac{1}{4}n^4$$
$$\frac{1}{4}\frac{n^4}{(n^2+1)^2}<\frac{\sum_{i=1}^ni^3}{(n^2+1)^2}<\frac{1}{4}\frac{4n^3+n^4}{(n^2+1)^2}$$
