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Use the non-increasing and non-decreasing theorem to show that ${S_n}$ converges. With $\beta>0$, $$S_n= \frac{\alpha+ n}{\beta + n}$$

(a) Taking $\alpha= 5$ and $\beta=6$ $$\frac{5+1}{6+1} \approx 0.85...$$ $$\frac{5+2}{6+2} \approx 0.875...$$ $$\frac{5+3}{6+3} \approx 0.888...$$ It shows that

$$S_n= \frac{\alpha+ n}{\beta + n} \leq \frac{\alpha+ n+1}{\beta + n+1} = S_{n+1}$$ It follows that ${S_n}$ is non-decreasing.

(b)

Quoting the theorem,

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

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

(c)

If $\epsilon >0 $, by definition of supremum, $$sup\{S_n\} - \epsilon< S_n \leq sup\{S_n\} $$

Also $$S_n \leq S_{n+1} \leq sup\{S_n\}$$

it follows that for every $\epsilon >0$ $$sup\{S_n\} - \epsilon < S_{n+1} \leq sup\{S_n\}$$ $$|S_{n+1} - sup\{S_n\}|< \epsilon$$

as $n+1 \geq n$

$$\lim\limits_{n \rightarrow \infty} S_{n+1} = sup\{S_n\}$$

It follows that $S_n$ converges.

The instruction is stating to show. I believe I did more a proof than a show case. Is what I did correct? How can this be shown using more directly maybe $\frac{\alpha +n}{\beta + n}$? How can I show more rigourously that $S_n$ is non-decreasing?

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  • $\begingroup$ Is $\alpha $ positive too? $\endgroup$ – randomgirl Aug 5 '17 at 9:34
  • $\begingroup$ Only $\beta>0$. nothing is mentioned about $\alpha$ $\endgroup$ – rei Aug 5 '17 at 9:36
  • $\begingroup$ what happens if $\alpha>\beta$? I think you should do cases. $\endgroup$ – randomgirl Aug 5 '17 at 9:38
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0) Case $\beta = \alpha$ $\rightarrow$ $S_n = 1$ , convergent.

Let $ \beta \ne \alpha$ :

Consider:

$f(x) = \frac{\alpha + x}{\beta +x}, x\in \mathbb{R^+}, \beta \gt 0$.

$f'(x) = $

$\frac{(\beta +x) - (\alpha +x)}{(\beta +x)^2} = \frac{(\beta - \alpha)}{(\beta +x)^2}$ ;

1) $\beta \gt \alpha \rightarrow$ $f'(x) \gt 0$.

2) $\beta \lt \alpha \rightarrow$ $f'(x) \lt 0$.

Choose $x = x_n = n$.

1) $x_{n+1} \gt x_n$ $\rightarrow$ $S_{n+1} \gt S_n $, increasing.

2) $x_{n+1} \gt x_n$ $\rightarrow$ $S_{n+1} \lt S_n $, decreasing.

Now find for 1) an upper bound, and for 2) a lower bound.

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I don't know what theorems you are using but I like this:

$f(n)=\frac{\alpha+n}{\beta+n}=\frac{\alpha-\beta}{\beta+n}+1$

Differentiating gives

$f'(n)=- \frac{\alpha-\beta}{(\beta+n)^2}$ Then determine when you will have increasing or decreasing... This will rely on the numerator $\beta-\alpha$

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    $\begingroup$ @rei ... you should see three different possible sequences here... an increasing one, a decreasing one, and one that is constant. $\endgroup$ – randomgirl Aug 5 '17 at 9:56
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HINT: use that $$S_{n+1}-S_n=\frac{\beta-\alpha}{(\beta+n+1)(\beta+n)}$$

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