2
$\begingroup$

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 $1/3^n < \varepsilon$?

$\endgroup$
  • 1
    $\begingroup$ just take $N$ such that $3^N > \frac{1}{\epsilon}$ $\endgroup$ – Jorge Fernández Hidalgo May 31 '17 at 18:21
  • $\begingroup$ 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. $\endgroup$ – Kaj Hansen May 31 '17 at 18:22
  • 2
    $\begingroup$ $$\frac1{3^n}<\varepsilon\iff\varepsilon^{-1}<3^n\iff n>-\log_3(\varepsilon)=\lfloor N\rfloor$$ $\endgroup$ – Simply Beautiful Art May 31 '17 at 18:27
2
$\begingroup$

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, $\lim_{i\rightarrow \infty} a_{i}=0$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Do you mean ratio test ? $\endgroup$ – Vivek Kaushik May 31 '17 at 19:02
  • $\begingroup$ I've already corrected. Thanks! $\endgroup$ – Vitor Alves May 31 '17 at 19:04
1
$\begingroup$

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 $0$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $n\to\infty$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.