MATH 3150 Class Notes 02-13-2025

notereal-analysis

Recall

For $L\in \mathbb{R}$, ${\lim }_{n\to \infty }{x}_n=L$ if $\forall \epsilon >0$, there exists $N>0$ so that $\left|x_n-L\right|<\epsilon$, for all $n>N$.

Monotone Convergence Theorem

Example

Let

$$\{\begin{array}{l}{x}_{n+1}=\sqrt{2+x_n},\\ x_1=\sqrt{2}.\end{array},n\ge 1.$$

Prove ${\lim }_{n\to \infty }{x}_n=2$.

Prove the Necessary Conditions are Satisfied:

Show $(x_n{)}_n$ is increasing.

Method 1:

1. $x_n\le x_{n+1}$ if and only if $\sqrt{2+x_{n-1}}\le \sqrt{2+x_n}$. Square both sides to get $2+x_{n-1}\le 2+x_n$, so $x_{n-1}\le x_n$.
2. Assume that $x_{n-1}\le x_n$, then we have $x_n\le x_{n+1}$, and since $\sqrt{2}\le \sqrt{2+\sqrt{2}}$, by induction we have that $(x_n{)}_n$ is increasing.

Method 2:

1. $x_n\le x_{n+1}$ if and only if $x_n\le \sqrt{2+x_n}$. Square both sides to get, $x_n^2\le 2+x_n$, which is equivalent to $x_n^2-x_n-2\le 0⟺(x_n+1)(x_n-2)\le 0⟺-1\le x_n\le 2$.
2. Since we already know that $x_n>0$, we just need to show that $x_n\le 2$. (Side note: this method may seem much more cumbersome, but we will need to show the sequence is bounded later, so this achieves the proof of both necessary conditions at the same time).
   1. $x_1=\sqrt{2}\le 2$.
   2. Assume $x_n\le 2$, then $x_{n+1}=\sqrt{2+x_n}\le \sqrt{2+2}=2$.
   3. Therefore, $x_n\le 2$.

Now we must show $(x_n{)}_n$ is bounded. By method 2 we’ve shown that the sequence is bounded ($0\le x_n\le 2$), if we proceeded by method 1, then we must now show that $(x_n{)}_n$ is bounded.

So far: $(x_n{)}_n$ is monotone and bounded, so by the monotone convergence theorem, we have that there exists $L\in \mathbb{R}$ such that ${\lim }_{n\to \infty }{x}_n=L$.

Why $L=2$: $x_{n+1}=\sqrt{2+x_n}⟹(x_{n+1}{)}^2=2+x_n$. Facts:

1. If ${\lim }_{n\to \infty }{x}_n=L⟹{\lim }_{n\to \infty }{x}_{n+1}$. Thus, $\left|x_n-L\right|<\epsilon \forall n>N⟹\left|x_{n+1}-L\right|<\epsilon \forall n>N-1$.
2. If ${\lim }_{n\to \infty }{x}_n=L_1$, and ${\lim }_{n\to \infty }{y}_n=L_2$, then ${\lim }_{n\to \infty }{x}_n{y}_n=L_1{L}_2$, so in particular, ${\lim }_{n\to \infty }(x_n{)}^2=L^2$.
   1. This will be proved next class.

$(x_{n+1}{)}^2=2+x_n$, so ${\lim }_{n\to \infty }(x_{n+1}{)}^2=2+{\lim }_{n\to \infty }{x}_n$, so $L^2=2+L$. We can solve this to get $L=-1$ and $L=2$, of which we can discard $L=-1$ because $L\ge 0$. $\square$

Infinite Limit

Example

${\lim }_{n\to \infty }\frac{n^4-5n^3}{n+\sqrt{n}+1}=\infty$.

We need to show that $\frac{n^4-5n^3}{n+\sqrt{n}+1}>M$, for all $n>N$.