# Does this product converge?

Does this product converge? $$\prod_{x=0}^{\infty}\frac{(30x+5)(30x+9)(30x+15)(30x+21)(30x+27)}{(30x+7)(30x+13)(30x+19)(30x+23)(30x+31)}$$

I have tried solving this by hand, and it seems to get closer to zero.

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The product is over $n$, but the terms do not depend on $n$. – JimmyK4542 Jul 24 '14 at 23:10
Is $x \in \mathbb{N}$ or $x \in \mathbb{R}$? – William Wu Jul 25 '14 at 1:56
Thanks for all the answers, btw @pomelo x∈N. – 7uc Jul 25 '14 at 5:44

## 2 Answers

$$P=\prod_{x=0}^{\infty}\frac{(30x+5)(30x+9)(30x+15)(30x+21)(30x+27)}{(30x+7)(30x+13)(30x+19)(30x+23)(30x+31)}$$

$$\frac{(30x+5)}{(30x+7)} <1$$

$$\frac{(30x+9)}{(30x+13)} <1$$

$$\vdots$$

$$P < 1$$

You are multiplying many numbers which are less than $1$ to the final result must be less than $1$.

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Further, the product is monotonically decreasing and positive; hence must have a limit point in $\mathbb{R}$. – vadim123 Jul 24 '14 at 23:21
Thank you @Brad for the answer. Sorry for sounding amateurish, but does that mean it converges to zero or just to a really small number? And if so, is there a way I could find that tiny number? – 7uc Jul 26 '14 at 21:03
All I showed is that the product is going to be less than $1$ which implies that it is bounded. The product does wind up converging to zero and that can be shown by proving that $$\prod_{x=0}^{\infty}\frac{30x+5}{30x+7}$$ converges to zero while the rest of the product is bounded. You can see the other answer for an idea how to show it goes to zero. – Brad Jul 26 '14 at 23:51
Oh! I understand now. Thank you! @Brad – 7uc Jul 27 '14 at 17:08

Take the natural log and use the fact that $\ln(1-y) \le -y$ for $y > 0$ to get

$\ln P_N = \displaystyle \sum_{x = 0}^{N}\left[\ln\dfrac{30x+5}{30x+7} + \ln\dfrac{30x+9}{30x+13} + \ln\dfrac{30x+15}{30x+19} + \ln\dfrac{30x+21}{30x+23} + \ln\dfrac{30x+27}{30x+31}\right]$

$= \displaystyle \sum_{x = 0}^{N}\ln\left[1-\tfrac{2}{30x+7}\right] + \ln\left[1-\tfrac{4}{30x+13}\right] + \ln\left[1-\tfrac{4}{30x+19}\right] + \ln\left[1-\tfrac{2}{30x+23}\right] + \ln\left[1-\tfrac{4}{30x+31}\right]$

$\le \displaystyle -\sum_{x = 0}^{N}\left[\dfrac{2}{30x+7} + \dfrac{4}{30x+13} + \dfrac{4}{30x+19} + \dfrac{2}{30x+23} + \dfrac{4}{30x+31}\right] \to -\infty$ as $N \to \infty$ by comparison to the harmonic series.

Since, $\ln P_N \to -\infty$, we have that $P_N \to 0$.

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