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I am not being able to find the specific product $\prod_{r=1}^{k} \left(1-\frac{1}{\sqrt {r+1}}\right)$ so to evaluate the given problem when $k \to \infty $.

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  • $\begingroup$ Presumably it "diverges to 0". $\endgroup$ – almagest May 22 '16 at 11:27
  • $\begingroup$ @almagest Yes you are right. But how? $\endgroup$ – StubbornAtom May 22 '16 at 11:28
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    $\begingroup$ Why not use the usual approach? Each term is less than $e^{-1/\sqrt{r+1}}$ and the corresponding series diverges. $\endgroup$ – almagest May 22 '16 at 11:32
  • $\begingroup$ @StubbornAtom One at least knows that the product is less than $1$, for every term is less than $1$. $\endgroup$ – Simply Beautiful Art May 22 '16 at 11:37
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The product is $0$ (therefore by the definition of a converging infinite product your product does not really converge). For every integer $n\geq 1$ we have \begin{equation*} 0 \,<\, \prod_{r=1}^n\left(1-\frac{1}{\sqrt{r+1}}\right) \,\leq\, \exp\left(-\sum_{r=1}^n\frac{1}{\sqrt{r+1}}\right)~. \end{equation*} The series $\sum_{r=1}^\infty 1/\sqrt{r+1}$ diverges, so the partial products of your infinite product tend to $0$.

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If you know that, for $0<a_n<1$, we have $$\prod_{i=1}^{\infty} (1-a_n) = 0 \text{ if and only if } \sum_{i=1}^{\infty} a_n = \infty$$ then it suffices if you show that $$\sum_{r=1}^\infty \frac1{\sqrt{r+1}} = +\infty.$$

For the proof of the above general fact, see How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$?

For the series which you get see, for example, Is $\sum\frac{1}{\sqrt{n+1}}$ convergent or divergent?

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