Solving Series Questions: Limit of $\frac{(2n-1)!}{(2n+1)!}$ & More

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The discussion focuses on solving limits of sequences and series, specifically the limit of the expression $\frac{(2n-1)!}{(2n+1)!}$, which simplifies to $\frac{1}{(2n+1)(2n)}$ and approaches 0 as n approaches infinity. Additionally, the sequence defined by $a_n = \sqrt{2 + a_{n-1}}$ is shown to be increasing and bounded above by 3, leading to a limit of 2. The participants emphasize the importance of understanding the definitions of series and sequences in tackling these problems.

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annie122
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I have some series related questions:

1) Find the limit of $\frac{(2n-1)!}{(2n+1)!}$

I know each term is less than one, but i don't know how to use this to get the limit

2) Find the limit of the sequence $\left(\sqrt{2},\,\sqrt{2\sqrt{2}},\,\sqrt{2\sqrt{2 \sqrt{2}}},\cdots\right)$

3) $a_n = \sqrt{2 + a_n-1}$

a) show $(a_n)$ is increasing and bounded above by 3.
b) Find the limit of $a_n$.

I have no clue for the last two.
 
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Note that

$$(2n+1)!= (2n+1) 2n (2n-1)!$$

The second question can be restated to find

$$a_{n+1}=\sqrt{2a_n }$$ where $a_0 =1$
 
Yuuki said:
I have some series related questions:
Also please note that "series" is the sum of the terms of a sequence.
 
Are you sure the 3rd question is stated correctly?

$$a_n=\sqrt{2+a_n-1}$$?
 
thanks, i now got the answer for the first one.
but i still don't know how to approach the problem even if it's restated that way.

and yes, I'm sure i wrote it as it is in the textbook.
 
Hint:

Try induction.
 

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