MHB Find Limit of $\frac{n^3}{(n + 1)^2}$ as $n$ Approaches ∞

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To find the limit of $\frac{n^3}{(n + 1)^2}$ as $n$ approaches infinity, the expression can be simplified to $\frac{n^3}{(n + 1)^2} = \frac{n^3}{n^2(1 + \frac{1}{n})^2}$. This leads to the form $\frac{n}{(1 + \frac{1}{n})^2}$, which is easier to analyze. As $n$ approaches infinity, $(1 + \frac{1}{n})^2$ approaches 1, simplifying the limit to $n$. Therefore, the limit diverges to infinity. The final result indicates that the limit does not converge to a finite value.
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I need to find the limit of

$$\left| \frac{(n + 1)n^3}{(n + 1)^{3}} \right|$$

as $n$ approaches infinity.

I simplify this to:

$$\left| \frac{n^3}{(n + 1)^{2}} \right|$$

But the solution simplifies it to:

$$\left| \frac{n}{(1 + \frac{1}{n})^{2}} \right|$$

How do I get to this result?
 
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tmt said:
I need to find the limit of

$$\left| \frac{(n + 1)n^3}{(n + 1)^{3}} \right|$$

as $n$ approaches infinity.

I simplify this to:

$$\left| \frac{n^3}{(n + 1)^{2}} \right|$$

But the solution simplifies it to:

$$\left| \frac{n}{(1 + \frac{1}{n})^{2}} \right|$$

How do I get to this result?
[math]\left | \frac{n^3}{(n + 1)^{2}} \right |[/math]

[math]= \left | \frac{n^2 \cdot n}{(n + 1)^{2}} \right | = \left | \frac{n}{\frac{(n + 1)^{2}}{n^2}} \right |[/math]

Can you finish from here?

-Dan
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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