# Trying to calculate a limit with a finite product and WolframAlpha disagrees with my logic.

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.

• 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 $\infty$. Dec 2, 2023 at 1:40

• 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 $$0$$: $$\frac{2^N N! (N+1)!}{(2N+1)!} \sim \frac{\sqrt{\pi N}}{2^{N+1}}.$$
• We used to say an infinite product evaluating to zero was divergent. Maybe your CAS reports divergence as $\infty$? Dec 2, 2023 at 0:31
• "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. Dec 2, 2023 at 20:53

Typing in Wolfram Alpha the command

FunctionExpand[Product[2j/(N+j+1),{j,1,N}]]

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

Typing now

Limit[(2^(-1-N)*Sqrt[Pi]*Gamma[2+N])/Gamma[3/2+N],N->Infinity]

it returns $$0$$ and the correct asymptotics.