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I'm trying to evaluate $$\lim_{n\to+\infty} \prod_{k=1}^{n} \frac{2k}{2k+1}$$ First I notice that since $k\geq1$ it is $\frac{2k}{2k+1}>0$ for all $k\in\{1,...,n\}$; so $$0\leq\lim_{n\to+\infty} \prod_{k=1}^{n} \frac{2k}{2k+1}$$ Then I notice that $$\prod_{k=1}^{n} \frac{2k}{2k+1}=\exp{\ln\left(\prod_{k=1}^{n} \frac{2k}{2k+1}\right)}=\exp{\sum_{k=1}^{n}\ln\left(\frac{2k}{2k+1}\right)}=$$ $$=\exp{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k+1}\right)}$$ Since $\ln(1+x)\leq x$ for all $x>-1$ and since $\exp$ is an increasing function it follows that $$\exp{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k+1}\right)}\leq\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}$$ So $$\lim_{n\to+\infty}\prod_{k=1}^{n} \frac{2k}{2k+1}\leq\lim_{n\to+\infty}\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}$$ Since $\exp$ is a continuous function it follows that $$\lim_{n\to+\infty}\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}=\exp{\sum_{k=1}^{+\infty}-\frac{1}{2k+1}}=e^{-\infty}=0$$ So by the comparison test we deduce that the limit is $0$.

Is this correct? Thanks for your time.

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    $\begingroup$ Yes your derivation is correct. +1 $\endgroup$ – Paramanand Singh Feb 9 at 14:13
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    $\begingroup$ I understand what $e^{-\infty}$ is supposed to represent, but I don't consider it particularly good style. I would write $$\exp \sum _{k=1}^n -\frac{1}{2k+1} \xrightarrow[n\to\infty]{}0 $$ But that's just a minor detail. +1 $\endgroup$ – Alvin Lepik Feb 9 at 14:24
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    $\begingroup$ In an infinite product like this, we say that it "diverges to zero". See math.stackexchange.com/q/3513417/442 $\endgroup$ – GEdgar Feb 9 at 14:40
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    $\begingroup$ Using the mathematical induction method it can be shown that the respective string is smaller than $\frac{1}{(n+1)^\frac{1}{2}}$. $\endgroup$ – medicu Feb 9 at 16:17
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Another way:

Using arithmetic geometric Inequality

$$\frac{k+k-1}{2}> \sqrt{k\cdot(k-1)}\Rightarrow \frac{2k-1}{2k}>\sqrt{\frac{k-1}{k}}$$

$$\frac{2k}{2k-1}<\sqrt{\frac{k-1}{k}}\Rightarrow \prod^{n+1}_{k=2}\frac{2k}{2k-1}<\prod^{n+1}_{k=2}\sqrt{\frac{k-1}{k}}=.\frac{1}{\sqrt{n+1}}$$

$$\Longrightarrow 0<\prod^{n+1}_{k=2}\frac{2k}{2k-1}<\frac{1}{\sqrt{n+1}}$$

Applying limit $n\rightarrow \infty$ and Using Squeeze Theorem

We have $$\prod^{n+1}_{k=2}\frac{2k}{2k-1}=0$$

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    $\begingroup$ The second line is incorrect: The first inequality should be $\frac{2k}{2k-1} < \sqrt{\frac{k}{k-1}}$. Consequently, several other statements are incorrect. $\endgroup$ – PolyaPal Mar 30 at 18:01
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Your caclulations are correct.

But I thought it might be interesting to see another nice trick.

  • Let $A_n = \prod_{k=1}^n\frac{2k}{2k+1}$ and $B_n = \prod_{k=1}^n\frac{2k+1}{2k+2}$.

Then, $A_n < B_n$ and $A_nB_n$ is a telescoping product and you get

$$A_n^2 < A_nB_n = \frac{2}{2n+2}=\frac 1{n+1}$$

Hence,

$$0 < A_n < \frac{1}{\sqrt{n+1}}\stackrel{n\to \infty}{\longrightarrow}0$$

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Yours is fine. Another simple method would be $$\lim_{n\to\infty}\prod_{k=1}^n\dfrac{2k}{2k+1}<\lim_{n\to\infty}\prod_{k=1}^n\dfrac{2k}{2k+2}\\ = \lim_{n\to\infty}\dfrac{1}{n+1}\\=0$$

EDIT

As Dr.WolfgangHintze pointed out in the comments, this inequality is in the reverse direction, so now we have the rate of decay of this series (from other answers). $$\dfrac{1}{\sqrt{n+1}}>\prod_{k=1}^n\dfrac{2k}{2k+1}>\dfrac{1}{n+1}$$

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  • $\begingroup$ @ Martund Looks nice, but, unfortunately, your inequality is wrong. $\endgroup$ – Dr. Wolfgang Hintze Feb 13 at 8:32
  • $\begingroup$ @Dr.WolfgangHintze, Why do you think that is wrong? $\endgroup$ – Martund Feb 13 at 18:24
  • $\begingroup$ Because $\frac{2k}{2k+1} > \frac{2k}{2k+2}$ and not $\lt$ $\endgroup$ – Dr. Wolfgang Hintze Feb 13 at 19:21
  • $\begingroup$ @ Martund I'm glad you could save your reasoning. Just a remark: for $n\ge2$ you can write your inequality even more elegant as $\frac{1}{\sqrt{n}}\gt \prod_{k=1}^n \frac{2k}{2k+1} \gt \frac{1}{n}$ $\endgroup$ – Dr. Wolfgang Hintze Feb 14 at 8:59
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Consider $$a_n=\prod_{k=1}^{n} \frac{2k}{2k+1}\implies \log(a_n)=\sum_{k=1}^{n}\log \left(\frac{2 k}{2 k+1}\right)=-\sum_{k=1}^{n}\log \left(1+\frac{1}{2 k}\right) $$ Use Taylor expansion $$\log \left(1+\frac{1}{2 k}\right)=\frac{1}{2 k}-\frac{1}{8 k^2}+O\left(\frac{1}{k^3}\right)$$ Compute the sum for this truncated series to get $$\log(a_n)\sim-\frac{H_n}{2}+\frac{H_n^{(2)}}{8}$$ where appear generalized harmonic numbers.

Use the asymptotics for large $n$ to get $$\log(a_n) =\left(\frac{\pi ^2}{48}-\frac{\gamma }{2}\right)-\frac{1}{2} \log \left({n}\right)-\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$

$$a_n\sim \frac{ \exp\left(\frac{\pi ^2}{48}-\frac{\gamma }{2}\right)}{\sqrt n}$$

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