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Show that $\prod_{n=1}^{\infty}(1+\frac{(-1)^n}{\sqrt{n}})$ is divergent.

Clearly the first term is 0, but the rest are positive real numbers. My textbook defines convergence even when a series includes a finite number of 0's, but Wikipedia (https://en.wikipedia.org/wiki/Infinite_product) states otherwise.

Furthurmore, Wolfram Alpha (https://www.wolframalpha.com/input?i2d=true&i=Product%5B1%2BDivide%5BPower%5B%5C%2840%29-1%5C%2841%29%2Cn%5D%2CSqrt%5Bn%5D%5D%2C%7Bn%2C2%2C%E2%88%9E%7D%5D) shows that the series starting from $n=2$ might converge to something.

To be more clear the definition I am using is: Let $p_n=\prod_{k=1}^{n}a_k$. If there exists some $N$ such that $a_n\neq 0$ for all $n>N$, then $\prod_{k=1}^{\infty}a_k$ converges if $\prod_{k=N+1}^{\infty}a_k$ converges (doesn't go to infinity or go to zero).

Is the question wrong, or does the series truly diverge?

Shean
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  • If we define an infinite product as the limit of the partial products, i.e.

    $$\prod_{i=1}^\infty a_i := \lim_{n \to \infty} \prod_{i=1}^n a_i$$

    then your product must evaluate to $0$, because $a_1 = 0$. There's no real ambiguity there.

     – PrincessEev Mar 22 '23 at 18:36
  • @PrincessEev I know it's zero. I've added the definition I am using for more clarity. – Shean Mar 22 '23 at 18:47
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    This duplicate answers your question: Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$., and your definition of the convergence of an infinite product is not the usual one. When according to you it converges to $0,$ it is usually said to diverge to $0.$ I found the duplicate using approach0 : https://approach0.xyz/search/?q=OR%20content%3A%24%5Cprod_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(1%2B%5Cfrac%7B(-1)%5En%7D%7B%5Csqrt%7Bn%7D%7D)%24&p=1 – Anne Bauval Mar 22 '23 at 18:55
  • @AnneBauval Thank you for that, not sure how to search for these types of questions. Google doesn't seem to give me the right questions. – Shean Mar 22 '23 at 18:58

1 Answers1

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Ignoring the first factor, it converges to $0$. (We may say diverges to $0$ as for infinite product, having a zero limit is considered to be a kind of divergence.)

Denote $a_n = 1 + \dfrac{(-1)^n}{\sqrt{n}}$, we can see that $a_{2 k - 1} \cdot a_{2 k} < 1 - \dfrac{1}{2 k - 1}$.

Then $\ln(a_{2 k - 1} \cdot a_{2 k}) < \ln \biggl( 1 - \dfrac{1}{2 k - 1} \biggr) < -\dfrac{1}{2 k - 1}$. Hence $$ \begin{aligned} \frac{1}{a_2} \prod_{i = 2}^{2 n} a_i = \prod_{i = 3}^{2 n} a_i &= \prod_{k = 2}^{n} a_{2 k - 1} \cdot a_{2 k} \\ &= \exp \ln \prod_{k = 2}^{n} a_{2 k - 1} \cdot a_{2 k} \\ &= \exp \sum_{k = 2}^{n} \ln(a_{2 k - 1} \cdot a_{2 k}) \\ &< \exp \sum_{k = 2}^{n} -\frac{1}{2 k - 1} \\ &= \exp\biggl( -\sum_{k = 2}^{n} \frac{1}{2 k - 1} \biggr) \\ &= 1 / \exp \sum_{k = 2}^{n} \frac{1}{2 k - 1} \text{.} \end{aligned} $$

You can see that this converges to $0$.

Moreover, you can use this to estimate the convergence rate. Note $H_{2 n} - H_n / 2 \sim (\ln 2 - 1/2) \ln n$, so the partial product has an order of $\Theta(n^{1 / 2 - \ln 2})$. (Hopefully my calculation is correct.)

Wolfram|Alpha gives a wrong answer is mainly because it used NProduct in Mathematica (which is not accurate).

(Sorry for answering a duplicate question, but my answer is independently written by myself and I think it has a better presence and more information.)

PinkRabbit
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