Why does a product $\lim_{n\to\infty} \prod_{i=1}^n (1- \frac{1}{n+2i})$ lead to infinity?

Question

I asked Wolfram Alpha to compute the following limit:

$\lim_{n\to\infty} \prod_{i=1}^n (1- \frac{1}{n+2i})$ and the answer was $\infty$. I do not understand how this can be, as all elements in the product are smaller than 1.

For comparison, when I asked about the following limit:

$\lim_{n\to\infty} \prod_{i=1}^n (1- \frac{1}{n+i})=1/2$.

What is the real value of these limits?

Answer 1

We may as well treat the more general product $$\prod_{i = 1}^n \left(1 - \frac{1}{n + a i}\right), \qquad a > 0 .$$ Rewriting gives $$\prod_{i = 1}^n \left(\frac{i+\frac{n - 1}a}{i + \frac{n}a}\right) = \frac{\prod_{i = 1}^n \left(i+\frac{n - 1}a\right)}{\prod_{i = 1}^n \left(i + \frac{n}a\right)} = \frac{\left(1 + \frac{n - 1}a\right)_n}{\left(1+\frac{n}a\right)_n},$$ where $(b)_n$ denotes the Pochhammer symbol. Rewriting this expression using the Gamma function representation $$(b)_n = \frac{\Gamma(n + b)}{\Gamma(n)}$$ of the Pochhammer symbol and using the Stirling series of $\Gamma$, $$\Gamma(s) = \sqrt{\frac{2 \pi} s} \left(\frac{s}e\right)^s \left(1 + \frac1{12 s} + R_1(s)\right), \quad R_1(s) \in O\left(\frac1{s^2}\right),$$ gives that as $n \to \infty$, the ratio behaves as $$\prod_{i = 1}^n \left(1 - \frac{1}{n + a i}\right) = (a + 1)^{-\frac1a} \left(1 + \frac{a - 1}{2 (a + 1) n} + R_2(n)\right),$$ where $R_2(n) \in O\left(\frac1{n^\frac32}\right)$. In particular, $$\boxed{\lim_{n \to \infty} \prod_{i = 1}^n \left(1 - \frac{1}{n + a i}\right) = (a + 1)^{-\frac1a}} .$$ For $a = 1$ the limit is $\boxed{\frac12}$, and for $a = 2$ the limit is $\boxed{\frac1{\sqrt3}}$

The same argument applies to the case $a > -1$, $a \neq 0$; in the limiting case $a = 0$ the product is $\left(1 - \frac1n\right)^n$, which has limit $\frac1e$ as $n \to \infty$.

I don't see how to explain WolframAlpha's behavior. Mathematica returns the correct result using the input Limit[Product[1 - 1/(n + 2*i), {i, 1, n}], n -> Infinity], but WolframAlpha gives $\infty$ for the same.

Answer 2

Here's WolframAlpha's incorrect reasoning as identified through the "Show steps" button.

$$\prod_{i=1}^n \left(1-\frac{1}{n+2i}\right)=\frac{\left(\frac{n+1}{2}\right)_n}{\left(\frac{n}{2}+1\right)_n}$$ (where that's the Pochhammer function)

By the quotient rule, we can distribute the limit operation over the fraction; the limit of the denominator is claimed incorrectly without further reasoning to be $1$, while the limit of the numerator is claimed without further reasoning to be $\infty$.

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You can get the right answer from WolframAlpha by taking logs:

Exp@Limit[Sum[Log[1 - 1/(n + 2 i)], {i, 1, n}], n -> \[Infinity]]

$\frac{1}{\sqrt{3}}$

Answer 3

Edit: Seems Patrick Stevens beat me to it

For what it's worth, here's a lightly-edited copy-paste of the step-by-step """solution""" Wolfram offers.

Find the following limit: $$ \lim_{n \to \infty} \prod_{i=1}^n \left(1 - \frac{1}{n+2i}\right).$$ Apply $\displaystyle\prod_{i=1}^n \left(1 - \frac{1}{n + 2i}\right) = \frac{\left(\frac{n+1}{2}\right)_n}{\left(\frac{n}{2} + 1\right)_n} $ to get $$\lim_{n\to\infty} \frac{(\frac{n+1}{2})_n}{(\frac{n}{2}+1)_n}.$$ Applying the quotient rule, write $$\lim_{n\to\infty} \frac{(\frac{n+1}{2})_n}{(\frac{n}{2}+1)_n} = \frac{\displaystyle\lim_{n\to\infty} (\frac{n+1}{2})_n}{\displaystyle\lim_{n\to\infty} (\frac{n}{2}+1)_n}.$$ Now $\displaystyle\lim_{n\to\infty} (\frac{n}{2} + 1)_n = 1$, so $$\lim_{n\to\infty} \left( \frac{n+1}{2}\right)_n = \infty$$

I guess the issue is occurring in the step where it asserts $$\lim_{n\to\infty} \left(\frac{n}{2} + 1\right)_n = 1.$$

Answer 4

Taking the logarithm of the product and applying a Taylor expansion we get $$ \log\left(\prod_{i=1}^n \left(1 - \frac{1}{n + ki}\right)\right)= \sum_{i=1}^n \log\left(1 - \frac{1}{n + ki}\right) =-(1+o(1)) \sum_{i=1}^n \frac{1}{n + ki} $$ as $n\to \infty$, where $k\in \mathbb{R}^+$ is fixed. The last term converges as a Riemann sum to $$ \int_1^n \frac{1}{n + kx} \, dx = \frac{\log(k+1)}{k}(1+o(1)), $$ as $n\to \infty$. Thus $$ \lim_{n \to \infty} \prod_{i=1}^n \left(1 - \frac{1}{n + ki}\right)=\exp\left({\lim\limits_{n\to \infty}\sum_{i=1}^{n}\log\left(1 - \frac{1}{n + ki}\right)}\right)=(k+1)^{-\frac{1}{k}}. $$