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Show that $\prod\limits_{i=1}^\infty\left(1-\frac{\alpha}{i}\right)=0$ if $\alpha > 0$.

Hint: Look at the logarithm of the absolute value of the product.

My attempt

If $\alpha\in\mathbb{N}$ then $\exists i=\alpha$ so one factor will equal $0$ and thus the desired result will be attained. I now consider the case when $\alpha\notin\mathbb{N}$. I don't know how to go about this so I attempt to use the hint.

$$\prod_{i=1}^\infty\left\vert1-\frac{\alpha}{i}\right\vert=\operatorname{exp}\left(\ln\left(\prod_{i=1}^\infty\left\vert1-\frac{\alpha}{i}\right\vert\right)\right) = \operatorname{exp}\left(\sum_{i=1}^\infty\ln\left(\left\vert1-\frac{\alpha}{i}\right\vert\right)\right)$$

I don't know where to go from there. I assume that I am to show that the series diverges to $-\infty$. I have briefly had a look at series but in the book I'm currently studying I haven't gotten to the series part yet so I don't think the author expects me to use a bunch of fancy methods to deduce the divergence of the series.

Michael Rozenberg
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David
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  • Related: http://math.stackexchange.com/q/2081399/147873 (This uses the Taylor series for $\log(1-x)$ + the fact that $\sum \frac{1}{n}$ diverges) – Winther Jan 09 '17 at 00:42
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    Hint: Use the upper bound $$1-\frac{\alpha}i\leqslant e^{-\alpha/i}$$ for every $i$, and the fact that $$\sum_i\frac{\alpha}i=+\infty$$ – Did Jan 09 '17 at 00:46
  • @Did I'm not familiar with that upper bound. Is it from some identity with a particular name I can look up? – David Jan 09 '17 at 00:53
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    @David You may derive that inequality just by looking at the positions of the graph of $y=e^{-x}$ and its tangent at $x=0$. – HorizonsMaths Jan 09 '17 at 01:14
  • @Did Thank you very much. – David Jan 09 '17 at 01:28

1 Answers1

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What you can do is split the infinite product into two parts, one part containing all of the values of $(1-a/i)$ whose absolute value is less than one, the other containing all the values of $(1-a/i)$ whose absolute value is greater than one (those two parts multiplied equal the original infinite product). The first part has an infinite amount of elements that decrease the value, so it equals zero (there is an infinite amount of elements in the first part because if $i$ is greater than $a$, $(1-i/a)$ is less than one, and there is an infinite amount of values for i that satisfy that). The second part contains a finite amount of elements that increase the value, so it equals a finite number, (there is a finite amount of elements because the absolute value of $(1-a/i)$ is greater than $1$ if $i$ is less than $a$, and because $i$ is positive $i$ is greater than zero and less than $a$, which is a finite gap). By multiplying the two parts you will get the original infinite product, the first part is zero and the second part is a finite number. A finite number times zero equals zero, therefore the original infinite product equals zero.

Vinyl_cape_jawa
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