11
$\begingroup$

Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.

I.) In the first step I studied monotony: $a_{n+1}-a_{n}=\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\cdot\cfrac{-2}{4n+7}<0$, $\{a_n\}$ is decreasing.

II.) In the second step I studied boundary. $$1>a_{1}=\cfrac{5}{7}>a_{2}=\cfrac{45}{77}>\dots>a_{n}>0$$

III.) In the last step I know that $\{a_n\}$ converges to $a\in\mathbb R$. $$a_{n+1}=a_{n}\cdot\cfrac{4n+1}{4n+3}$$ Taking the limit as $n\to\infty$: $$a=a$$

No conclusion.

But if I apply Cesaro-Stolz?

IV.) Let $\{x_n\}_{n\ge1}^{\infty}=\{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)\}$ and $\{y_n\}_{n\ge1}^{\infty}=\{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)\}$. Then $$\lim_{n\to \infty}{\cfrac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=\lim_{n\to \infty}{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)^2}{7\cdot11\cdot15\cdot\dots.\cdot(4n+5)}=?}}$$

If you have a simple solution, I would appreciate it. Thank you!

$\endgroup$

3 Answers 3

10
$\begingroup$

One can check that $$ \bigg( \frac{4n+1}{4n+3} \bigg)^2 < \frac{n+1}{n+2} $$ for all $n\ge0$. Therefore $$ 0 < a_n = \frac57 \frac9{11} \cdots \frac{4n+1}{4n+3} < \bigg( \frac23 \frac34 \cdots \frac{n+1}{n+2} \bigg)^{1/2} = \sqrt{\frac2{n+2}}, $$ and so $a_n\to0$ by the squeeze theorem.

$\endgroup$
2
  • $\begingroup$ Nice and simple. I was going to add something simpler to my answer, but it's not needed now (+1) $\endgroup$
    – robjohn
    Jan 15, 2014 at 18:40
  • $\begingroup$ That's ok, I appreciated the pointer to Gautschi's Inequality. $\endgroup$ Jan 15, 2014 at 23:51
4
$\begingroup$

Using the relation $x\Gamma(x)=\Gamma(x+1)$, we get $$ \frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac74\right)}\frac{\frac54\frac94\frac{13}4\cdots\frac{4n+1}4}{\frac74\frac{11}4\frac{15}4\cdots\frac{4n+3}4} =\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)} $$ Therefore, $$ \frac{5\cdot9\cdot13\cdots(4n+1)}{7\cdot11\cdot15\cdots(4n+3)} =\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\frac{\Gamma\left(\frac74\right)}{\Gamma\left(\frac54\right)} $$ Using Gautschi's Inequality, we get $$ \left(\frac{4n+7}4\right)^{-1/2}\le\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\le\left(\frac{4n+3}4\right)^{-1/2} $$ By the Squeeze Theorem, we get that the limit is $0$.

$\endgroup$
1
$\begingroup$

We know that infinite products of the form $\displaystyle\prod_k(1+a_k)$ have the same nature as infinite series of the form $\displaystyle\sum_ka_k$ . Here, $a_k = \frac{4k+1}{4k+3}-1 = -\frac2{4k+3}$, which has order of magnitude $\frac1k$. Therefore the product diverges, having the same nature as the harmonic series. Also, an infinite product is said to diverge to $0$, which is clearly the case here, since each term is lesser than $1$.

$\endgroup$
1
  • $\begingroup$ Cool - I'll delete my comments that are no longer relevant. I added a sentence to your answer - feel free to keep or delete it as you wish. $\endgroup$ Jan 16, 2014 at 4:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .