Find the value of $\lim_{n \rightarrow \infty} \Big( 1-\frac{1}{\sqrt 2} \Big) \cdots \Big(1-\frac{1}{\sqrt {n+1}} \Big)$

Question

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

Attempt:

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.

Same question: https://math.stackexchange.com/questions/1795186/how-do-i-evaluate-prod-r-1-infty-left-1-frac1-sqrt-r1-right. — StubbornAtom, Apr 17 '17 at 10:45

score 8 · Accepted Answer · answered Apr 30 '15 at 11:38

8

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 $n\to+\infty$?

answered Apr 30 '15 at 11:38

mickep

19,962

The limit tends to $0$. – MathMan Apr 30 '15 at 11:46
That is indeed correct. – mickep Apr 30 '15 at 11:47
2

The limit is $0$, or the sequence tends to $0$. – GEdgar Apr 30 '15 at 12:28
@GEdgar, that is indeed correct. (I agree with anyone commenting this answer.) – mickep Apr 30 '15 at 12:37

Find the value of $\lim_{n \rightarrow \infty} \Big( 1-\frac{1}{\sqrt 2} \Big) \cdots \Big(1-\frac{1}{\sqrt {n+1}} \Big)$

1 Answers1

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