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

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}

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    $\begingroup$ (small comment on language, monotony means something like lack of variety, the word you meant to write is monotonicity) $\endgroup$ Oct 27 '18 at 18:27
  • $\begingroup$ Thank you, I didnt knew that because it's not my native language. $\endgroup$
    – Doesbaddel
    Oct 27 '18 at 18:29
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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

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  • $\begingroup$ That helped a lot, thank you. Unfortunately I can't choose 2 answers as accepted. $\endgroup$
    – Doesbaddel
    Oct 31 '18 at 12:56
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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 $\lim_{n \to \infty}a_n=1$.

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  • $\begingroup$ Thank you! I need to show monotony for all $n\in \mathbb{N}$ that's why the sequence is not monotone (and not divergent?) for all $n$ although it sure is for $n\geq 3$ right? $\endgroup$
    – Doesbaddel
    Oct 27 '18 at 18:12
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    $\begingroup$ @Doesbaddel that is why I have used the phrase ultimately monotonic. For convergence etc. the behavior of finitely many terms is irrelevant. $\endgroup$
    – Anurag A
    Oct 27 '18 at 18:33
  • $\begingroup$ Oh I didnt knew that. Did I get this right: You only need to have a infinite monotonic part of the sequence after a finite number of terms $a_n$ with $n\in N$ that are not monotonic in order that you can say it's convergent? $\endgroup$
    – Doesbaddel
    Oct 27 '18 at 18:45
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    $\begingroup$ @Doesbaddel Infinite monotonic part is a bit confusing. Because you may have something like $1,\frac{1}{2}, 2,\frac{1}{3},3, ...$ in which case it has infinite monotonic part but the sequence is NOT convergent. That is why we use the word ultimate which conveys that barring some finite terms, the behavior goes on from some point onwards. Now even ultimate monotonicity is NOT a guarantee for convergence. For example, $1,3,1/2,2,3,4,5,6, \ldots$ is ultimately monotonic but divergent. Combination of ultimate monotonicity and appropriate boundedness (above or below) gives convergence. $\endgroup$
    – Anurag A
    Oct 27 '18 at 18:51
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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).

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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.

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