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

Question

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.

Answer 1

• 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 $0$: $$ \frac{2^N N! (N+1)!}{(2N+1)!} \sim \frac{\sqrt{\pi N}}{2^{N+1}}. $$

Answer 2

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.