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Prove that the sequence $(a_n)$ converges, where $a_n$ = $\frac{1 · 3 · 5 · · ·(2n − 1)}{2 · 4 · 6 · · ·(2n)}$. Use Monotonic Convergence Theorem.

Here is my proof:

Since $a_n = \frac{2n-1}{2n}$, then $a_{n+1} = \frac{2n-1+1}{2n+1} = \frac{2n}{2n+1}$. Therefore, $a_n$ = $\frac{2n-1}{2n}\le\frac{2n}{2n+1}$ = $a_{n+1}$, and so the sequence, $a_n$, is increasing for all $n\in\mathbb{N}$, and is monotonic. The term $a_n$ has an upper bound (ie. 1) and a lower bound (ie. 0) for all $n$ and thus is bounded. Therefore, $a_n$ converges by the Monotonic Convergence Theorem.

Please check it over and make sure it sounds all right. If there is anything wrong, please let me know what and how I can fix it.

vitamin d
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red
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    How do you get from $a_n = \frac{1 · 3 · 5 \cdots (2n − 1)}{2 · 4 · 6 \cdots(2n)}$ to $a_n = \frac{2n-1}{2n}$? – Martin R Mar 15 '21 at 17:19
  • While $\frac{2n-1}{2n}$ is an increasing sequence, its partial products are a decreasing sequence with lower limit $0$. Do you want to prove $\lim_{n\to\infty}a_n$ exists, or $\lim_{n\to\infty}\sum_{k=1}^na_k$ exists? – J.G. Mar 15 '21 at 17:19
  • See https://math.stackexchange.com/q/1025630/42969 for a correct solution. – Martin R Mar 15 '21 at 17:21

2 Answers2

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You can start by noting that the sequence is decreasing. In fact, $$ \dfrac{a_{n+1}}{a_n} = \dfrac{2n+1}{2n+2} < 1 \Rightarrow a_{n+1}< a_n. $$

Also, the sequence is clearly bounded. Since it is decreasing and positive, we have that $0 < a_n \leq a_1$.

Finally, being monotonic and bounded, the sequence is convergent.

PierreCarre
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  • I'm sorry, I'm really new to proofs. Can you explain how you got 2n+2 in the denominator? – red Mar 15 '21 at 17:35
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    $$\frac{a_{n+1}}{a_n} = \dfrac{\dfrac{1\cdot 3 \cdot \cdots \cdot (2n-1)(2n+1)}{2 \cdot 4 \cdot \cdots \cdot (2n)(2n+2)}}{\dfrac{1\cdot 3 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot \cdots \cdot (2n)}}$$

    So, everuthing cancels out, except for $\frac{2n+1}{2n+2}$.

     – PierreCarre Mar 15 '21 at 17:39
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Without looking at the dup solution, you can find a recursive relation for $a_{n+1}$ and $a_n$. One has: $a_{n+1} = \dfrac{1\cdot 3\cdot 5\cdots (2n+1)}{2\cdot 4\cdot 6\cdots (2n+2)}= \dfrac{2n+1}{2n+2}\cdot a_n< a_n$ since $2n+1 < 2n+2$. And $a_n > 0$ for all $n$, hence it converges as a monotonically decreasing and bounded below sequence.