null sequence

Null sequence

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence such that it converges to 0:

$$\underset{n\to \infty }{\lim }{a}_n=0$$

Then we call $(a_n{)}_{n\in \mathbb{N}}$ a null sequence.

${\left(\frac{1}{n}\right)}_{n\in \mathbb{N}}$, $(2^{-n}{)}_{n\in \mathbb{N}}$ are null sequences.

$(a_n)$ converges to $a$ if and only if $(a_n-a)$ is a null sequence.

$$a_n\to a⟺(a_n-a)\to 0$$

${\left(1+\frac{1}{n}\right)}_{n\in \mathbb{N}}$ is not a null sequence, but converges to 1.

Comparison lemma

If $(a_n)$ is a null sequence and $|b_n|\le C\cdot |a_n{|}^c$ for some $c,C>0$ and almost all $n$, then $(b_n)$ is also a null sequence.

The $C$ and $c$ generalize the obvious case $|b_n|\le |a_n|$: constant multiples and powers of null sequences are still null.

Corollary: $\left(\frac{C}{n^c}\right)$ is null for any $c,C>0$
Apply the lemma with $a_n=\frac{1}{n}$. Then $C\cdot |a_n{|}^c=C\cdot \frac{1}{n^c}=\frac{C}{n^c}\to 0$.

Common null sequences

Geometric decay: $z^n\to 0$ for $|z|<1$. (Diverges for $|z|>1$; for $|z|=1$ it depends on the argument.)

Polynomial decay: $\frac{1}{n^k}\to 0$ for any $k>0$.

Exponential beats polynomial: $\frac{n^k}{z^n}\to 0$ for $|z|>1$ and any $k\in \mathbb{Z}$.

Proof that $z^n\to 0$ for $|z|<1$

For $z=0$ it’s trivial. For $z\ne 0$, write $|z|=\frac{1}{1+x}$ with $x>0$.
By the bernoulli inequality, $(1+x{)}^n\ge 1+nx$, so:

$$|z^n|=\frac{1}{(1+x{)}^n}\le \frac{1}{1+nx}\le \frac{1}{nx}=\frac{1}{x}\cdot \frac{1}{n}$$

Since $\frac{1}{n}\to 0$, we have $z^n\to 0$.

Proof that $\sqrt[n]{n}\to 1$

Let $x_n=\sqrt[n]{n}-1$, so $n=(1+x_n{)}^n$. We show $x_n\to 0$.

Since $n\ge 1$, we have $x_n\ge 0$. Using the binomial expansion:

$$n=(1+x_n{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2})x_n^2=\frac{n(n-1)}{2}{x}_n^2$$

Rearranging: $x_n^2\le \frac{2}{n-1}$, so $|x_n|\le \sqrt{\frac{2}{n-1}}$.

Since $\frac{1}{\sqrt{n}}\to 0$, we get $x_n\to 0$, i.e. $\sqrt[n]{n}\to 1$.

$\sqrt[n]{a}\to 1$ for any $a>0$

$$\sqrt[n]{a}=a^{\frac{1}{n}}\to a^0=1$$

$n$	$\sqrt[n]{n}$
2	$\sqrt{2}\approx 1.41$	$1.41\times 1.41=2$
10	$\approx 1.26$	$1.26^{10}=10$
100	$\approx 1.047$	$1.047^{100}=100$
1000	$\approx 1.007$	$1.007^{1000}=1000$