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Find the bound of the following sequences:

$$a)\ \frac{n^2}{n^2+1}$$ $$b)\ \frac{n^3}{n+1}$$

My attempt:

$a)$$$0<\frac{n^2}{n^2+1}<1, \forall n\in \Bbb N $$ $$\frac{n^2}{n^2+1}>0 \Leftrightarrow (n^2>0) \land (n^2+1)>0$$ which is true $\forall n\in \Bbb N$.

$$\frac{n^2}{n^2+1}<1 \Leftrightarrow \frac{n^2}{n^2+1}-1<0 \Leftrightarrow \frac{-1}{n^2+1}<0 \Leftrightarrow n^2>-1$$

which is also true $\forall n\in\Bbb N$.

Is this correct?

Similarly I managed to show that in $b)\ \frac{n^3}{n+1}>0, \forall n\in\Bbb N$ but I don't know what to do about the upper bound. I assume it doesn't exist, but how do I prove it?

  • 0
    HINT: Write $n^3 = (n^2-n+1)(n+1)-1$. Another way is to use $n+1 \le 2n$ for $n \ge 1$. – 2017-01-11
  • 0
    A little detail: you don't **find** the boundedness, you **prove** or **establish** it. – 2017-01-11

3 Answers 3

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$a)$ is correct, and so are the bounds.

For $b)$, it is true that $\forall n \in \mathbb N$, both $n^3 \geq 0$ and $n+1 \geq 0$ (infact, $n+1 \geq 1$) hold true. Therefore, their quotient is also positive i.e. $\frac {n^3}{n+1} \geq 0$ is also true.

However, to analyse the other direction, we can use the following trick: $$ \frac{n^3}{n+1} = \frac{n^3+1}{n+1} - \frac 1{n+1} = n^2-n+1 - \frac{1}{n+1} $$ Now, note that :$\frac 1{n+1} \leq 1$, so $-\frac 1{n+1} \geq -1$. Finally, we can say that: $$ n^2-n+1 - \frac{1}{n+1} \geq n^2-n+1 -1 \geq n^2-n \geq n(n-1)\geq n-1 $$

Therefore, as $n$ becomes larger, $n-1$ also becomes infinitely large. But then, we have prove that $n-1 \leq \frac{n^3}{n+1} $, so it follows that $\frac{n^3}{n+1} $ is unbounded.

There are many tricks to show these kinds of limits, I have shown you one.

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Let $a_n = \frac{n^2}{n^2+1}$ be the first sequence from your question.

A sequence $c_n$ is bounded if for all $n \in \mathbb{N}$, $$M_1 \leq c_n \leq M_2,$$ that is, bounded from below by $M_1$ and bounded from above by $M_2$.

In order to prove a statement $a_n \leq 1$, try to be a little more specific, like: $$a_n = \frac{n^2}{n^2+1} < \frac{n^2+1}{n^2+1} = 1,$$ and $$a_n = \frac{n^2}{n^2+1} > \frac{0}{n^2+1} = 0.$$ That way we have found $M_1 = 0$ and $M_2 = 1$. Can you establish a lower bound for $b_n$? And a (possibly) upper bound for $b_n$ and why?

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for $n\ge 1$ you have $\frac{1}{2} \leq \frac{n}{n+1}$ indeed $1+\frac{1}{n}\leq 2$ that is $\frac{1}{1+\frac{1}{n}}\geq\frac{1}{2}$. Then, $\frac{n^2}{2}\leq\frac{n^3}{n+1}$. and clearly $\frac{n^2}{2}$ in unbouded