# $(1-\frac{1}{n})^m$ tends to zero for all $n \in \mathbb{N}$ fixed when $m$ tends to infinity?

Regards

I want to proof that for all $n \in \mathbb{N}$ fixed $$\lim_{m \rightarrow \infty} \left( 1-\frac{1}{n} \right)^m = 0$$

Is the last statement true?

• Yes, because $0 \leq 1-\frac{1}{n} < 1$ Mar 5, 2014 at 22:08
• Yes, it's true. More generally, if $0 \leq a < 1$ then $a^n \to 0$ as $n \to \infty$. Mar 5, 2014 at 22:08

Yes is a consequence of the following claim:

If $|r|<1$ then $r^n \to 0$.

Define $\{|r^n|: n\in \mathbb{N}\}$, the set is bounded below by zero, let $a= \inf\{|r^n|: n\in \mathbb{N}\}$ note that $a\ge 0$, we shall show that $a=0$. Suppose that $a\not=0$. Then $a/|r|>a$ so $a/|r|>|r|^n$, thus $a>|r|^{n+1}$, i.e., $a$ cannot be the greatest lower bound, a contradiction. Thus $a=0$.

Given $\varepsilon>0$, choose $N$ such that $|r|^N<\varepsilon$, since $|r|<1$ for all $n\ge N$, we have $|r^n|\le |r^N|< \varepsilon$, i.e., $|r^n|<\varepsilon$ for all $n\ge N$. Hence $r^n \to 0$.

• I don't understand how you get $a/|r| > |r|^n$ from $a/|r| > a$ Mar 5, 2014 at 23:11
• @MarioLamasEspinoza : $a/|r|>a$ means that $a/|r|$ cannot be the greatest lower bound because is greater than this, and this is not possible. Then there is some member of the set such that $|r|^n<a/|r|$ Mar 5, 2014 at 23:19
• Ok. You mean that there is a $N \in \mathbb{N}$ such that for all $n \geq N$ we have $a/|r| > |r|^n$ Mar 5, 2014 at 23:28
• @MarioLamasEspinoza : Yes there is some element which is less than $a/|r|$ and from here get the contradiction and conclude that the $\inf$ is $0$. To show that this hold for all $n\ge N$, $|r^n|=|r|^N |r|^{n-N}\le |r|^N < \varepsilon$. This is because $n-N\ge 0$ and $|r|<1$. =) Mar 6, 2014 at 0:58

Yes true. It's well known that for a geometric sequence we have

$$a^n\xrightarrow[n\to\infty] \,0\quad \iff |a|<1$$

Proof:

• The case $a=0$ is clear.
• Let $a\ne0$ then we have $$n\log(|a|)\xrightarrow[n\to\infty] \,-\infty$$ since $\log(|a|)<0$ so we deduce the result: if $|a|<1$ then $\displaystyle\lim_{n}a^n=0$.
• But how to prove it? Are logs known by then? Take logs, so that $\log a < 0$ and $\log(a^n) = n(\log a)$ goes to $-\infty$. Mar 5, 2014 at 22:19
• I'm just writing the proof.
– user63181
Mar 5, 2014 at 22:22

The sequence $$\left( 1-\frac{1}{n} \right)^m$$ for fixed $n$ and $m\in \mathbb N$ is strictly decreasing with $m$ because $\left( 1-\frac{1}{n} \right)\lt 1$

Every term of the sequence positive so the sequence is bounded below by zero. Therefore it has a greatest lower bound.

Suppose this is $L\gt 0$, and let $M=\cfrac L{1-\frac 1n}\gt L$. Then there is a term of the sequence less than $M$, and then the next term of the sequence is less than $L$.

Some points to refine, but that should do it.