# Show that the sequence $(1/3^n)$ converges.

I can see that the limit is 0, but I don't know how to prove convergence. From the definition of convergence, for $\varepsilon > 0$ There exists a natural number N such that for all $n \geq N$, $|1/3^n-0| < \varepsilon$. So how do I prove $1/3^n < \varepsilon$?

• just take $N$ such that $3^N > \frac{1}{\epsilon}$ – Jorge Fernández Hidalgo May 31 '17 at 18:21
• Notice that we can make $3^n$ arbitrarily large. How large does it need to be to guarantee that $1/3^n < \varepsilon$? So suppose we know we must have $3^n>M$. Take a natural log of both sides to figure out how large $n$ needs to be. – Kaj Hansen May 31 '17 at 18:22
• $$\frac1{3^n}<\varepsilon\iff\varepsilon^{-1}<3^n\iff n>-\log_3(\varepsilon)=\lfloor N\rfloor$$ – Simply Beautiful Art May 31 '17 at 18:27

Consider the serie $S=\sum\limits _{i=1}^{\infty} \frac{1}{3^{i}}$. By ratio test,we have to $\lim\limits_{i \rightarrow \infty} \frac{a_{i+1}}{a_{i}}=\frac{1}{3} <1$. Hence, $S$ converges. Therefore, $\lim_{i\rightarrow \infty} a_{i}=0$.

• Do you mean ratio test ? – Vivek Kaushik May 31 '17 at 19:02
• I've already corrected. Thanks! – Vitor Alves May 31 '17 at 19:04

Let $1 < r = 1 + s \implies s \ge 0$

Using the binomial theorem, $r^n = (1+s)^n = 1 + ns + \frac{n(n-1)}2s^2 > 1 + ns > ns$

Taking the reciprocal.

$\frac1{r^n} = \frac1{(1+s)^n} < \frac{1}{1+ns} < \frac1{ns}$

To show convergence (to $0$), we need to find $N$ s.t. $\forall \epsilon > 0$, if $n>N$ then

$|\frac{1}{r^n}| < |\frac1{ns}| < \epsilon$

Which is satisfied for $N = \frac{1}{s\epsilon}$.

If we let $r = 3$, we have proved your sequence converges to $0$

### Statement: For a given $\epsilon >0\space \exists$ a $N\in \mathbb{N}$ such that $N\epsilon >1$

Proof: Suppose not. Then for all $N\in \mathbb{N}$, $N\epsilon \leq 1$, then $N\leq 1/\epsilon \Rightarrow N< ([1/\epsilon]+1)=N_0$ (Here $[x]$ is the maximum integer which is less than or equals to $x$.), but by our assumption the inequality holds for every $N\in \mathbb{N}$, then does not exists any $N_0$ i.e. is a contradiction.

Now you can easily show that $1/3^n < \epsilon$ for any $\epsilon >0$ as $n\to\infty$.