Limit Of A Sequence Examples 1

Limit of a Sequence Examples 1

Example 1

Prove that $\lim_{n \to \infty} \left \{ \frac{1}{n} \right \} = \infty$.

Let $\epsilon > 0$ be given. We want to find an integer $N$ such that if $n > N > 0$ and $n \in \mathbb{N}$ then $\biggr \rvert \frac{1}{n} - 0 \biggr \rvert < \epsilon$.

We first start with the following inequality noting that both $n > 0$ and $N > 0$.

(1)
\begin{align} \biggr \rvert \frac{1}{n} \biggr \rvert < \epsilon \\ \frac{1}{n} < \epsilon \\ n > \frac{1}{\epsilon} \end{align}

If we choose $N = \left \lceil \frac{1}{\epsilon} \right \rceil$ (We note that $\left \lceil x \right \rceil$ is equal to the smallest integer that is greater or equal to $x$. We include this step to ensure that $N \in \mathbb{N}$), then if $n > N = \frac{1}{\epsilon}$, then:

(2)
\begin{align} \biggr \rvert \frac{1}{n} \biggr \rvert = \frac{1}{n} < \frac{1}{N} = \frac{1}{\frac{1}{\epsilon}} < \epsilon \end{align}

Therefore $\lim_{n \to \infty} \left \{ \frac{1}{n} \right \} = 0$.

Example 2

Prove that $\lim_{n \to \infty} \frac{n + 1}{n} = 1$.

Let $\epsilon > 0$ be given and find $N \in \mathbb{N}$ such that if $n ≥ N$ then $\biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert < \epsilon$.

Let's first do some algebraic manipulation with the following inequality:

(3)
\begin{align} \quad \biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert = \biggr \rvert \frac{n+1}{n} - \frac{n}{n} \biggr \rvert = \biggr \rvert \frac{n+1 - n}{n} \biggr \rvert = \biggr \rvert \frac{1}{n} \biggr \rvert \end{align}

But we know that since $n \in \mathbb{N}$ ($n$ is a natural number), then $n > 0$, so we can get rid of the absolute value bars, that is:

(4)
\begin{align} \biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert = ... = \biggr \rvert \frac{1}{n} \biggr \rvert = \frac{1}{n} < \epsilon \end{align}

Therefore, let's look at the inequality $\frac{1}{n} < \epsilon$ which implies that $n > \frac{1}{\epsilon}$. We note that if we choose $N > \frac{1}{\epsilon}$, then our inequalities will hold. Let's show it:

(5)
\begin{align} N > \frac{1}{\epsilon} \\ n ≥ N > \frac{1}{\epsilon} \\ \frac{1}{n} ≤ \frac{1}{N} < \epsilon \\ \quad \biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert = \frac{1}{n} ≤ \frac{1}{N} < \epsilon \implies \biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert < \epsilon \end{align}

Therefore we have shown that $\lim_{n \to \infty} \frac{n + 1}{n} = 1$.

Example 3

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

We first let $\epsilon > 0$ be given. We want to find an integer $N$ such that if $n > N > 0$ and $n \in \mathbb{N}$ then $\mid f(x) - L \mid < \epsilon$.

We will first start off with the following inequality:

(6)
\begin{align} \biggr \rvert \frac{2n^2}{n^2 + 1} - 2 \biggr \rvert < \epsilon \\ \biggr \rvert \frac{2n^2 - 2(n^2 + 1)}{n^2 + 1} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{2n^2 - 2n^2 + -2}{n^2 + 1}\biggr \rvert < \epsilon \\ \biggr \rvert \frac{-2}{n^2 + 1} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{2}{n^2 + 1} \ \biggr \rvert < \epsilon \\ \frac{2}{n^2 + 1} < \epsilon \\ \frac{n^2 + 1}{2} >\frac{1}{\epsilon} \\ n^2 + 1 > \frac{2}{\epsilon} \\ \end{align}
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