This question already has an answer here:

$$\lim_{n \rightarrow \infty} \Big( 1-\dfrac{1}{\sqrt 2} \Big) \cdots \Big(1-\dfrac{1}{\sqrt {n+1}} \Big)$$


Let $y = \lim_{n \rightarrow \infty} \Big( 1-\dfrac{1}{\sqrt 2} \Big) \cdots \Big(1-\dfrac{1}{\sqrt {n+1}} \Big)$

$\ln y = \lim_{n \rightarrow \infty} \ln \Big( 1-\dfrac{1}{\sqrt 2} \Big)+ \cdots + \ln \Big(1-\dfrac{1}{\sqrt {n+1}} \Big)$

I am not able to move ahead really from here. Could someone give me an hint on how to move forward with this problem.

Thank you very much for your help in this regard.


marked as duplicate by Sil, mrtaurho, Delta-u, ancientmathematician, max_zorn Feb 24 at 21:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


All the factors are positive, but bounded by the last one. Thus, for all $n$ the product is between $0$ and $(1-1/\sqrt{n+1})^n$. What happens with that expression as $n\to+\infty$?

  • $\begingroup$ The limit tends to $0$. $\endgroup$ – MathMan Apr 30 '15 at 11:46
  • $\begingroup$ That is indeed correct. $\endgroup$ – mickep Apr 30 '15 at 11:47
  • 2
    $\begingroup$ The limit is $0$, or the sequence tends to $0$. $\endgroup$ – GEdgar Apr 30 '15 at 12:28
  • $\begingroup$ @GEdgar, that is indeed correct. (I agree with anyone commenting this answer.) $\endgroup$ – mickep Apr 30 '15 at 12:37

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