I want to calculate $$\lim_{N \to \infty} \prod_{j=1}^{N} \frac{2j}{N + j + 1} \,.$$ Here is my logic:

  • For the range of the product operator ($1 \leq j \leq N $) the inequality $2j < N + j + 1$ is always true. Therefore $\frac{2j}{N + j + 1} < 1 $, and so an upper boundary can be estabilished using the largest term of the product: $$\lim_{N \to \infty} \prod_{j=1}^{N} \frac{2j}{N + j + 1} \leq \lim_{N \to \infty}\frac{2N}{2N + 1} = 1 \,.$$

However, when I use WolframAlpha I get $\infty$ for an answer.

I have seen other questions that challenge the reliability of WolframAlpha, so I am tempted to dismiss it.

My questions:

  • Is my logic correct or is WolframAlpha correct? (and why?)
  • If the limit is indeed finite, does it go to zero or does it simply coneverge to some value smaller than 1? How can I show it?

Thank you very much.

  • 1
    $\begingroup$ I use Limit[Product[2*j/(N + j + 1), {j, 1, N}], N -> Infinity] in Mathematica, it is zero. Use the same code in www.wolframalpha.com, it is $\infty$. $\endgroup$
    – River Li
    Dec 2, 2023 at 1:40

2 Answers 2

  • You're right and WolframAlpha is wrong.
  • Note that the first factor is $\frac2{N+2}$ and all the other factors (as you noted) are less than $1$. So the product is actually less than $\frac2{N+2}$, which shows that the limit equals $0$. (For what it's worth, when the expression is copied exactly from WolframAlpha into Mathematica, it correctly evaluates the limit as $0$.)
  • The product is exactly equal to $\frac{2^N N! (N+1)!}{(2N+1)!}$. (WolframAlpha can correctly evaluate the limit as $0$ when the expression is written in this form.) A standard application of Stirling's formula gives the rate of convergence to $0$: $$ \frac{2^N N! (N+1)!}{(2N+1)!} \sim \frac{\sqrt{\pi N}}{2^{N+1}}. $$
  • 2
    $\begingroup$ We used to say an infinite product evaluating to zero was divergent. Maybe your CAS reports divergence as $\infty$? $\endgroup$ Dec 2, 2023 at 0:31
  • $\begingroup$ "WolframAlpha can correctly evaluate the limit as 0 when the expreession is written in this form" So, then there must be some issue with calculating limits with Pochhammer symbols. $\endgroup$
    – rus9384
    Dec 2, 2023 at 20:53

Typing in Wolfram Alpha the command


it properly returns $$\frac{\sqrt{\pi }\,\, 2^{-N-1}\,\, \Gamma (N+2)}{\Gamma \left(N+\frac{3}{2}\right)}$$

Typing now


it returns $0$ and the correct asymptotics.


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