Calculating $\prod_{x=1}^{\infty}\frac{x}{x+1}$ I am trying to find the value of: $$\prod_{x=1}^{\infty}\frac{x}{x+1}$$
It is equal to $\frac12\times\frac23\times\frac34\times\ldots$ and we can observe that all terms are cancelling out. So it converges to zero. but I don't know how can I proceed algebraically without writing the terms and multiplying them. for example I wrote it as $$\dfrac{\prod_{x=1}^{\infty}x}{\prod_{x=1}^{\infty}(x+1)}$$
But not sure how to continue.
 A: You are on the right track. Use the partial product:
$$\prod_{x=1}^\infty \frac{x}{x+1} = \lim_{R\to\infty} \prod_{x=1}^R \frac{x}{x+1}= \lim_{R\to\infty} \frac{\prod_{x=1}^R x}{\prod_{x=1}^R(x+1)} = \lim_{R\to\infty} \frac{1}{R+1} = 0.  $$
A: By definition,$$\require{cancel}\prod_{x=1}^\infty\frac x{x+1}=\lim_{n\to\infty}\prod_{x=1}^n\frac x{x+1}.$$But, for each $n\in\Bbb N$,$$\prod_{x=1}^n\frac x{x+1}=\frac1{\cancel2}\times\frac{\cancel2}{\cancel3}\times\cdots\times\frac{\cancel n}{n+1}=\frac1{n+1}.$$
A: State:  $\prod_{x=1}^n \frac x{x+1}= \frac 1{n+1}$.
Then either prove it by induction:
Base case: $\prod_{x=1}^1 \frac x{x+1}= \frac 12$.  Induction step: If $\prod_{x=1}^n\frac x{x+1}= \frac 1{n+1}$ then $\prod_{x=1}^{n+1} \frac x{+1} = (\prod_{x=1}^n\frac x{x+1}) \cdot \frac {n+1}{n+2} = \frac 1{n+1}\frac {n+1}{n+2} = \frac 1{n+2}$.
Or just state "it follows by induction" (it's clear enough that you don't have to go into details).
So $\prod_{x=1}^{\infty} = \lim_{n\to \infty} \prod_{x=1}^n \frac x{x+1}=\lim_{n\to \infty} \frac 1{n+1} = 0$.
