Is there an accurate way to estimate the product $\prod_{n=1}^{1009}\frac{2n+1}{2n}$?

Question

I don't have a background in mathematics, I just was doing this for fun...so if there are any flaws in any logic below please feel free to correct the logic.

Anyways, there was a question posed in this video (https://www.youtube.com/watch?v=jZPFIQ8kIUs) asking if $(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{6})...(1+\frac{1}{1009}) > 50$ or not.

The algebra behind why the product is less than $50$ is clear, but I was curious if there was a way to estimate or accurately determine what this number actually is or might be.

My attempt to find that out goes as follows:

$$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{6})...(1+\frac{1}{2018}) = \frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}\cdots\frac{2019}{2018} = \prod_{n=1}^{1009}\frac{2n+1}{2n}.$$

Doing a little bit of research I read that you can write a product $\prod a_i$ as: $$e^{\sum\ln(a_i)},$$ which is essentially a Riemann sum (I believe) and thus allows one to write: $$e^{\int \ln(a_i)}.$$

If the reason why $e^{\sum \ln(a_i)} = e^{\int \ln(a_i)}$ isn't because the summation isn't a Riemann sum, I would be curious to read a brief explanation in a comment below as to the reasoning for why they're equal.

Back to the calculation...

\begin{align} \prod_{n=1}^{1009}\frac{2n+1}{2n} = \prod_{n=1}^{1009}{(2n+1)}\cdot\prod_{n=1}^{1009}{\frac{1}{2n}} = e^{\int_1^n{\ln(2n+1)}}\cdot e^{\int_1^n{\ln(\frac{1}{2n})}} \quad (1) \end{align}

After a bit more research I figured out I could use Stirling approximation to determine the values of $\ln\big(\frac{1}{2n}\big)$ and $\ln(2n+1)$.

From the Stirling approximation Wikipedia page I learned that $\displaystyle \sum_{j=1}^n{\ln(j)} \approx \int_1^n {\ln(x) dx} = n\cdot \ln(n) - n + 1$, so: $$\sum_{j=1}^n{\ln(\frac{1}{2n})} = -2\cdot \sum_{j=1}^n{\ln(n)} = -2\cdot \int_1^n {\ln(x) dx} = -2\cdot (n\cdot \ln(n) - n + 1)\\and\\\sum_{j=1}^n{\ln({2n+1})} = \sum_{j=2}^{n+1}{\ln(2n)} = 2\cdot \sum_{j=2}^{n+1}{\ln(n)} = 2\cdot \int_2^{n+1} {\ln(x) dx} = 2\cdot (n\cdot \ln(n) - n + 1)$$So from $(1)$ we now have,$$e^{2\cdot \int_2^{n+1} {\ln(x) dx}}\cdot e^{-2\cdot \int_1^n {\ln(x) dx}} = \frac{e^{2\cdot \int_2^{n+1} {\ln(x) dx}}}{e^{2\cdot \int_1^n {\ln(x) dx}}}$$and with $n = 1009$, we have $$\frac{e^{\int_2^{1010} {\ln(x) dx}}}{e^{\int_1^{1009} {\ln(x) dx}}} \approx \frac{e^{5977.88266...}}{e^{5970.96545...}} = e^{6.91721} = 1009.49955 > 50$$

But we know that the product we're trying to estimate is less than $50$. In order to get a value for $x$ such that $e^{x}<50$, then $x < 3.912$. From the Stirling approximation, we found $x$ to be $6.91721$. However, upon further reading about Stirling approximation of $\ln(x)$ can be off by $0.008$. For the given value of $5977.88266...$ this means the real value can fall within $5977.88266... \pm 107$, which means the difference between $5977.88266...$ and $5970.96545...$ are within margin of error.

Despite them being so close to each other, to get the proper estimate, their difference needs to be less than $3.912$. However I don't know how to move forward in being able to find an estimation, unless the logic behind my estimation is flawed...in which case the solution would be found as per finding out the error.

But if my logic was sound and this discrepancy between $e^{6.91721}$ and $e^{x<3.912}$ is just the result of approximation errors in the Stirling approximation, then my question becomes:

Is there a way to better accurately determine or approximate this answer?

Answer 1

For the original question with the upper limit of the product $2018$ I would write $$\prod_{n=1}^{2018}\frac{2n+1}{2n}=\frac {4037!!}{4036!!}=\frac {4037!}{(4036!!)^2}=\frac {4037!}{2^{4036}(2018!)^2}$$ and it is time for Stirling's approximation $$\approx \frac {4037^{4037}e^{4036}}{{2^{4036}}2018^{4036}e^{4037}}\sqrt{\frac{4037}{2\pi 2018^2}}=4037\left(1+\frac 1{4036}\right)^{4036}\frac 1e\sqrt{\frac{4037}{2\pi 2018^2}}$$ The power on the parenthesized terms is very close to $e$, so this is about $$4037\sqrt{\frac{4037}{2\pi 2018^2}}\approx 4037(2018 \pi)^{-1/2}$$ and a hand calculation is enough to show that $2018 \pi \lt 6400 = 80^2$ and we know the value is over $50$. The final expression is about $50.7018$ compared to the Alpha value of $50.6986$ so our approximations have been very good. The same approach will work with $1009$ as the upper limit. We would get $$2019 \sqrt {\frac{2019}{2 \pi 1009^2}}\approx 35.869$$

Answer 2

The product you are trying to evaluate does not match the product given in the original problem, because the upper index is inconsistent between the two.

Specifically, you wrote

$$\prod_{n=1}^{\color{red}{2018}} \left(1 + \frac{1}{2n}\right) = \prod_{n=1}^{2018} \frac{2n+1}{2n} = \frac{3}{2} \cdot \frac{5}{4} \cdots \color{red}{\frac{4037}{4036}}$$ but this is not equal to the original product, which is

$$\prod_{n=1}^{\color{blue}{1009}} \left(1 + \frac{1}{2n}\right).$$ You must have $1009$ as the upper index if the final factor in the product is to take on the form $$\frac{2019}{2018}.$$

Answer 3

You can use the asymptotics for the ratio of two gamma functions: $$ \prod\limits_{n = 1}^N {\frac{{2n + 1}}{{2n}}} = \frac{2}{{\sqrt \pi }}\frac{{\Gamma (N + 3/2)}}{{\Gamma (N + 1)}} \sim 2\sqrt {\frac{N}{\pi }} \left( {1 + \frac{3}{{8N}} - \frac{7}{{128N^2 }} + \ldots } \right). $$ In your case $N=1009$. With the first three terms as given, we obtain the approximation $35.8560$. A simple two-sided inequality is as follows: $$ \frac{{2N + 1}}{{\sqrt {\pi \left( {N + \frac{4}{\pi } - 1} \right)} }} \le \prod\limits_{n = 1}^N {\frac{{2n + 1}}{{2n}}} \le \frac{{2N + 1}}{{\sqrt {\pi \left( {N + \frac{1}{4}} \right)} }} $$ for any $N\geq 1$. The constants $\frac{4}{\pi } - 1$ and $\frac{1}{4}$ are the best possible.

Answer 4

Let $P(m)=\prod_{n=1}^m{\frac{2n+1}{2n}}$. Using the Wallis product, $$\lim_{m\to\infty}\frac{2m+1}{(P(m))^2}=\frac{\pi}{2}$$

$$\therefore P(1009)\approx\sqrt{\frac{4(1009)+2}{\pi}}\approx 35.9$$

Answer 5

In the same spirit as @Gary, Stirling approximation gives the series $$\prod\limits_{n = 1}^k {\frac{{2n + 1}}{{2n}}} = \frac{2}{{\sqrt \pi }}\frac{ \Gamma \left(k+\frac{3}{2}\right)}{ \Gamma (k+1)}\sim\frac{2 }{\sqrt{\pi }}\sqrt{k}\,\sum_{p=0}^\infty \frac {a_p}{k^p}$$ Since $a_0=1$, we can replace the sum by the $[q,q]$ Padé approximant $P_q$. For example, $$P_2=\frac{1280 k^2-4368 k-2625}{1280 k^2-4848 k-737}$$ whose error is $O\!\left(\frac{1}{k^5}\right)$.

For $k=1009$ this gives $\color{red}{35.856013020145820}65$ while the exact value is $\color{red}{35.85601302014582018}$