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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$ ....

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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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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 $n \ge N$.

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