Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging

Question

$$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ $(0\in\mathbb{N})$

Monotonicity:

To prove, that a sequence is monotonic, I can use the following inequalities: \begin{align} a_n \leq a_{n+1}; a_n < a_{n+1}\\ a_n \geq a_{n+1}; a_n > a_{n+1} \end{align} I inserted some $n$'s to get an idea on how the sequence is going to look like.

I got: \begin{align} a_0&=3\\ a_1&=1\\ a_2&=\frac{7}{9}\approx 0.\overline{7}\\ a_3&=\frac{3}{4}=0.75 \end{align} Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$

Therefore, I show that \begin{align} a_n \leq a_{n+1}; a_n < a_{n+1}\\ a_n \geq a_{n+1}; a_n > a_{n+1} \end{align} I am having problems when trying to prove the inequalities above: \begin{align} & a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\ & = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\ & = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\ & = \cdots \text{ not sure what steps I could do now} \end{align}

Boundedness:

The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$: \begin{align} a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\ &=3 \end{align}

The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally.

Convergence

Assumption (s.a) $\lim_{ n \to \infty} a_n =1$

Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:

\begin{align} \mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\ &= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\ &= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\ &= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\ &= \left|\frac{2-2n}{(n+1)^2}\right|\\ &= \cdots \text{(how to go on?)} \end{align}

Answer 1

For monotonic behavior:

$a_{n+1}-a_n=\frac{(n+1)^2+3}{(n+2)^2}-\frac{n^2+3}{(n+1)^2}=\frac{(n+1)^4+3(n+1)^2-n^2(n+2)^2-3(n+2)^2}{(n^2+3n+2)^2}=\frac{2(n^2-n-4)}{(n^2+3n+2)^2}.$

Observe that for $n \geq 3$, the numerator is always positive, so for all $n \geq 3$, the sequence will be an increasing sequence (so ultimately an increasing sequence). The non-monotonic behavior only occurs in the first few terms $a_0 > a_1 > a_2>a_3<a_4<a_5<\dotsb<a_n<a_{n+1}<\dotsb$

For boundedness

\begin{align*} a_n & = \frac{n^2+3}{(n+1)^2}\\ & \leq \frac{n^2+3}{n^2}\\ & \leq 1+\frac{3}{n^2}\\ & \leq 4 & (\forall n \geq 1) \end{align*} This is also satisfied by $a_0=3$.

Since it is ultimately monotonic and bounded, hence convergent.

For convergence:

You already have \begin{align*} |a_n-1| & = \left|\frac{2-2n}{(n+1)^2}\right|\\ & \leq \frac{2n}{(n+1)^2}\\ & \leq \frac{2n}{n^2}\\ & \leq \frac{2}{n}. \end{align*}

For an $\epsilon >0$. Let $N$ be the smallest integer bigger than $\frac{2}{\epsilon}$. Then for all $n \geq N$, we have $$|a_n-1| \leq \epsilon.$$ This shows that $\lim_{n \to \infty}a_n=1$.

Answer 2

Hint: $$a_{n+1}-a_n=2\,{\frac {{n}^{2}-n-4}{ \left( n+2 \right) ^{2} \left( n+1 \right) ^{ 2}}} $$ Second hint: $$a_n=\frac{n^2(1+\frac{3}{n^2})}{n^2(1+\frac{2}{n}+\frac{1}{n^2})}$$ this tends to $1$ for $n$ tends to infinity

Answer 3

An alternative for monotonicity:

Consider the corresponding (continuous) function $f(x) = \dfrac{x^2 + 3}{(x+1)^2}$. You can then check (using calc 1 methods) that $f'(x) > 0$ if $x \geq 3$. Hence $a_n$ is increasing for $n \geq 3$. You have already checked that $a_1 \leq a_2 \leq a_3$, so the sequence is monotone increasing.

Alternatively, to show boundedness, first show that $a_n$ converges to $1$, and then use the fact that every convergent sequence is bounded (you should think about why this last fact is true).

Answer 4

In order to analyse the sequence, $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ We can look at the function, $$f(x) =\frac {x^2+3}{(x+1)^2}$$

$$f'(x) = \frac { x^2-2x-3}{(x+1)^4} >0 \text { for x>3 }$$ Therefore the sequence is increasing for $n>3$

Note that $$\lim _{x\to \infty } f(x) =1$$ Thus $$\lim _{n\to \infty } a_n=1$$

Since every convergent sequence is bounded, so is our sequence $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ is bounded. For example we can see $|a_n|<4$ for all positive integers.