How to prove the limit of a sequence using "$\epsilon-N$"

Question

I think I have a proper understanding of the general procedure, but I'm having difficulty manipulating my inequality so that I can isolate $n$ by itself. Sadly I wasn't given many examples to model my answer on.

Prove that $\displaystyle\lim_{n\to\infty}\frac{n+1}{n^2+1}=0$

So I'm given $L=0$. I then look at the inequality

$$\left| \frac{n+1}{n^2+1}-0\right|<\epsilon$$

but I have no idea how to isolate $n$. The best I can come up with, which may be the right idea, is to use another function $f$ such that

$$\left|\frac{n+1}{n^2+1}\right|<f<\epsilon$$

and then work with that. But my idea of using $f=\lvert n+1\rvert$ seems to have me a bit stuck too.

Answer 1

Hint: Show that $0<\frac{n+1}{n^2+1}<\frac{1}{n-1}$.

Answer 2

Since the given sequence is positive for all $n\geq 1$, we can drop the absolute value signs.

Consider the inequality: \begin{equation*} \frac{n+1}{n^2+1}<\epsilon \end{equation*} This is a messy inequality, as solving for $n$ would be rather difficult. Instead we shall find an upper bound for the numerator $(n+1)$ and a lower bound on the denominator $(n^2+1)$ so that we can construct a new sequence $f_{n}$ such that $$\frac{n+1}{n^{2}+1}<f_{n}<\epsilon.$$

Since $n+1$ behaves like $n$ at very large numbers, we shall try to find a value for $b$ such that $bn>n+1$. And since $n^{2}+1$ behaves like $n^{2}$ for very large numbers, we shall try to find a value for $c$ such that $cn^{2}<n^{2}+1$. $n+1<n+n=2n$ and $n^{2}+1>\frac{1}{2}n^{2}$ for all $n\geq 1$.

Now, consider the new inequality: \begin{align*} \frac{n+1}{n^{2}+1}<\frac{2n}{\frac{1}{2}n^{2}}<\frac{4}{n}<\epsilon \end{align*} We have \begin{equation*} \frac{4}{n}<\epsilon\iff \frac{4}{\epsilon}<n \end{equation*}

Let $\displaystyle N(\epsilon)=\lfloor4/\epsilon\rfloor$. Therefore, $\forall \epsilon>0,\ \exists N(\epsilon)=\lfloor4/\epsilon\rfloor\ni\forall n>N(\epsilon)$ \begin{align*} \frac{n+1}{n^{2}+1}&<\frac{4}{n}<\frac{4}{N(\epsilon)+1}<\frac{4}{4/\epsilon}=\epsilon \end{align*}

Answer 3

Let $\epsilon <1$ then $$\frac{n+1}{n^2+1}<\epsilon \Leftrightarrow \epsilon n^2-n-(1-\epsilon)>0$$

You can solve this inequeality like a quadratic polinomial and you can find that for $$\forall n\geqslant n_0=[\frac{1+2\sqrt{\epsilon(1-\epsilon)}}{2}]+1 \Rightarrow \frac{n+1}{n^2+1}<\epsilon$$ forall $\epsilon<1$.

Thus we found an appropraite $n_0 \in \mathbb{N}$,which depends on $\epsilon$, such that $\frac{n+1}{n^2+1}<\epsilon,\forall n\geqslant n_0$

This is a way to prove it but indeed it is easier to find a simpler sequence $b_n$ such that $\frac{n+1}{n^2+1} \leqslant b_n<\epsilon$.

Answer 4

The $\epsilon-N$ definition to $\lim_{n\to \infty}a_n=L$ is $$ \big( \forall \epsilon>0 \big) \big( \exists N_{\epsilon}\in \mathbb{N} \big) \big[ ( \forall n\in \mathbb{N} ) ( n>N_{\epsilon} ) \implies ( |a_n-L|<\epsilon ) \big] $$ That is, given an arbitrary, but fixed, $\epsilon>0$, we must find a candidate for $N_\epsilon\in\mathbb{N}$ with the property that $$ (\forall n\in \mathbb{N} ) ( n>N_{\epsilon} ) \implies ( |a_n-L|<\epsilon ) $$ The number $N_\epsilon$ also depends on the limit $L$ and the sequence itself as well. In this case, $L=0$ and $$ a_n= \frac{n+1}{n^2+1}. $$ To find an $N_\epsilon$ candidate with the required property we must manipulate the inequality $$ 0-\epsilon< \frac{n+1}{n^2+1}< 0 +\epsilon. $$ The inequality of the left is valid for all $\epsilon> 0$. Focus on the inequality of the right $$ \frac{n+1}{n^2+1}< \epsilon. $$ Note that $ \frac{n+1}{n^2+1} < \frac{n+1}{n^2-1} = \frac{1}{n-1} $ for $n\geq 2$. Once $\frac{1}{n-1}<\epsilon$ implies $\frac{n+1}{n^2+1}<\epsilon$, set $$ N_\epsilon =\min\left\{n\in\mathbb{N}: n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2 \right\} $$ In fact, \begin{align} n>N_\epsilon \implies & n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2 \\ \implies & n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2 \\ \implies & \frac{1}{\epsilon}<n-1 \mbox{ and } n\geq 2 \\ \implies & n-1>\frac{1}{\epsilon} \mbox{ and } n\geq 2 \\ \implies & \frac{1}{n-1}<\epsilon \mbox{ and } n\geq 2 \\ \implies & \frac{n+1}{n+1}\cdot\frac{1}{n-1}<\epsilon \\ \implies & \frac{n+1}{n^2-1}<\epsilon \\ \implies & \frac{n+1}{n^2+1}<\epsilon \\ \implies & -\epsilon< \frac{n+1}{n^2+1}< \epsilon. \\ \implies & \left|\frac{n+1}{n^2+1}\right|< \epsilon. \end{align}

Answer 5

Examining $\frac{n+1}{n^2+1}$ you realize that as $n$ grows it will start 'acting like' $1/n$. You know the quadratic formula, and after a bit of tinkering you guess that the key is analyzing

$\tag 1 \frac{n+1}{n^2+1} \le \frac{2}{n}$

The inequality (1) is equivalent to

$\tag 2 n^2 - n + 2 \ge 0$

The discriminant for this quadratic expression is negative, so it is always true.

Let the $\varepsilon \gt 0$ challenge be presented. Then $2/n \lt \varepsilon$ iff $n \gt 2/\varepsilon$.

Select any natural number $N$ greater than $2/\varepsilon$.

We have shown that if $n \ge N$, $\frac{n+1}{n^2+1} \lt \varepsilon$, and the proof of convergence had been demonstrated.

Answer 6

Fairly crude bounds or estimates can be very useful in simplifying problems about limits. For example if $n\geq 1$ then $$0<\frac {n+1}{n^2+1}=\frac {n(1+n^{-1})}{n^2(1+n^{-2})}=\left(\frac {1}{n}\right)\cdot \frac {1+n^{-1}}{1+n^{-2}}\leq \left( \frac {1}{n}\right) \cdot\frac {2}{1+n^{-2}}< \left(\frac {1}{n}\right)\cdot\frac {2}{1}=\frac {2}{n}.$$ So if $\epsilon >0$ and $n\geq \max (1,\frac {2}{\epsilon})$ then $|\frac {n+1}{n^2+1}|<\epsilon.$