Using the non-increasing/ non decreasing theorem , show that $\{S_n\} $ converges $$S_n= \frac{r^n}{1+r^n} $$ With $r>0$

The theorem to apply:

if $\{S_n\}$ is non-decreasing , then $\lim\limits_{n \rightarrow \infty} S_n = sup\{S_n\}$.

If $\{S_n\}$ is non-increasing, then $\lim\limits_{n \rightarrow \infty} S_n = inf\{S_n\}$

Nonincreasing/decreasing? Considering $$ S_{n+1} = \frac {r^{n+1} } {1+r^{n+1} }$$ $$ S_{n+1} = \frac {r \cdot r^n } {r \cdot (\frac{1}{r} +r^n}) $$ $$ S_{n+1} = \frac { r^n } {\frac{1}{r} +r^n } $$

When $r \geq 1$ $$1 \geq \frac{1}{r}$$ $$1+ r^n \geq \frac{1}{r} + r^n$$ $$S_n = \frac{r^n}{1+ r^n} \leq \frac{r^n}{ \frac{1}{r} + r^n } = S_{n+1} $$ It shows that the sequence is non decreasing by definition

It follows that by the theorem stating that if $S_n$ is non-decreasing, we have $$ \lim\limits_{n \rightarrow \infty} S_n = sup\{S_n\}$$

Considering that $$ S_n= \frac{r^n}{1+r^n} < \frac{r^n}{r^n} = 1 => sup\{S_n\}=1 = \lim\limits_{n \rightarrow \infty} S_n$$

For every $\epsilon > 0$, there exists an integer $N$ such that $| S_n – 1| < \epsilon$ with $n \geq N$

Considering $$ | S_n – 1| = |\frac{r^n}{1+r^n} -1 |= |- \frac{1}{1+r^n} | = \frac{1}{1+r^n} < \frac{1}{r^n}<\epsilon$$ Let N be such that $$ \frac{1}{r^N} < \epsilon$$ $$\frac{1}{\epsilon} < r^N$$ $$\log \frac{1}{\epsilon} < N \cdot \log r$$ $$N > - \frac{\log \epsilon}{log r}$$ For every $\epsilon >0$, there is $n\geq N > - \frac{\log \epsilon}{\log r}$ s.t. $$|S_n -1| < \epsilon$$ It follows that when $r \geq 1$, $S_n$ converges. .

Then there is the case when $ 0 < r \leq 1$ ....


I am unsure about those results. My question is about $N > - \frac{\log \epsilon}{\log r}$ and $ n\geq N > - \frac{\log \epsilon}{\log r}$ Is my method correct? can this be negative? I think I have some difficulties with the conceptual understanding when framing N. Does $N$ have to be a value or is a bound enough?

  • $\begingroup$ Your algebra will become easier if you rewrite as $S_n=\frac{1}{1+r^{-n}}=(1+r^{-n})^{-1}$. $\endgroup$ – vadim123 Aug 6 '17 at 5:19
  • 2
    $\begingroup$ You need to watch out for the case when $r=1$. Then $S_n$ is always $\frac{1}{2}$ so clearly convergent, but not to 1. Distinguish the cases $r>1$, $r=1$ and $0<r<1$. $\endgroup$ – Epiousios Aug 6 '17 at 6:20
  • $\begingroup$ Thx for this. I did not see that one. So It should have three cases: $r>1$, $r=1$, and $0<r<1$ $\endgroup$ – rei Aug 6 '17 at 6:24
  • 1
    $\begingroup$ @1524 Thanks for the edit to my post. You are absolutely correct that I intended for $r>1$ though I appear to have slipped up and did not write this. $\endgroup$ – Brevan Ellefsen Aug 6 '17 at 6:28

Everything looks correct to me but could be condensed and reworded a bit. Here is how I would do it:

Let $S_n = \frac{r^n}{1+r^n}=\frac{1}{1+r^{-n}}$ so that $S_n$ is clearly non-decreasing for $r\ge 1$ since $r^{-n}$ is non-increasing
Then $\sup(S_n)=\lim\left( \frac{1}{1+r^{-n}}\right) =1$
We now wish to prove that $S_n$ converges to $1$ when $r>1$; to this end, let $\epsilon > 0$ be given, and let $N\in\mathbb{N}$ such that $N> \frac{\log(1/\epsilon)}{\log(r)} \implies \frac{1}{r^N} < \epsilon$
Then, for all natural numbers $n \ge N$, $$|S_n - 1| = \left|\frac{1}{1+r^{-n}} - 1\right| = \left|\frac{-r^{-n}}{1+r^{-n}}\right| = \frac{r^{-n}}{1+r^{-n}}=\frac{1}{1+r^n}<\frac{1}{r^n} <\frac{1}{r^N}<\epsilon$$ And so we conclude that $S_n$ converges to $1$ .

Is my method correct?

As far as I can tell, yes it is.

Edit: as @1524 notes,this argument only works for $r > 1$ and not for $r \ge 1$. At the point $r=1$ we just have a sequence with all terms $\frac{1}{2}$ which trivially converges to $\frac{1}{2}$

can this be negative?

Note that $\frac{-\log(\epsilon)}{\log(r)} = \frac{\log(1/\epsilon)}{\log(r)}$ is only negative $\epsilon > 1$ in which case any positive value for $N$ will suffice.

Does $N$ have to be a value or is a bound enough?

$N$ is definitely a value, since $N \in \mathbb{N}$. It is also a lower bound on $n$, since $n \ge N$.


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.