Evaluate $\lim_{n \to \infty} n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)$

Question

Evaluate $\lim_{n \to \infty} n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)$

Attempt $1$:

$$1 - \frac1m + \frac5{4m ^ 2}=\frac{4m^2-4m+5}{4m^2}=\frac{(2m-1)^2+4}{4m^2}=\left(1-\frac1{2m}\right)^2+\frac1{m^2}$$

Don't know what to do with it.

Attempt $2$:

Trying to write the terms of the product to see if there is any pattern.

$$\prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)=\frac54\cdot\frac{13}{16}\cdot\frac{29}{36}\cdot\frac{53}{64}\cdots$$

The numerator terms have differences $8,16,24$...

Don't know what to do with it.

Answer 1

We have the Weierstrass product for $\cos$:

$$\cos(\pi z) = \prod_{m=1}^\infty\left(1-\frac{4z^2}{(2m-1)^2}\right)$$ valid for all $z\in \Bbb C$.

Setting $z=i$ we get $$\cosh(\pi) = \prod_{m=1}^\infty\left(1+\frac{4}{(2m-1)^2}\right)$$

We also have the Wallis product:

$$\prod_{m=1}^\infty\frac{(2m)^2}{(2m-1)(2m+1)} = \frac{\pi}{2}$$

Then

\begin{align} n\prod_{m=1}^n\left(1-\frac{1}{m}+\frac{5}{4m^2}\right)&=n\prod_{m=1}^n\left(\frac{(2m-1)^2+4}{(2m)^2}\right)\\ &=n\prod_{m=1}^n\left(\frac{(2m-1)^2+4}{(2m-1)^2}\right)\prod_{m=1}^n\left(\frac{(2m-1)^2}{(2m)^2}\right)\\ &=\frac{n}{2n+1}\prod_{m=1}^n\left(1+\frac{4}{(2m-1)^2}\right)\prod_{m=1}^n\left(\frac{(2m-1)(2m+1)}{(2m)^2}\right)\\ &\underset{n\to\infty}\longrightarrow \frac{1}{2}\cosh(\pi)\frac{2}{\pi} = \frac{\cosh(\pi)}{\pi}\end{align}

Answer 2

This is only a partial answer which couldn't quite fit as a comment, but it at least shows that the limit does exist and is not $0$.
We have: $$\begin{split} \ln\left(1 - \frac{1}{m} + \frac{5}{4m ^ 2}\right) &= - \left(\left(\frac{1}{m} - \frac{5}{4m ^ 2}\right) + \frac{1}{2} \left(\frac{1}{m} - \frac{5}{4m ^ 2}\right)^2 + o\left(\left(\frac{1}{m} - \frac{5}{4m ^ 2}\right)^2\right)\right)\\ &= - \frac{1}{m} + \frac{1}{4m^2} + \frac{\varepsilon_m}{m^2}\end{split}$$ for some sequence $(\varepsilon_m)_m$ converging to $0$.
Thus, making the $\frac{1}{m}$ go to the other side and after summing from $m = 1$ to $n$: $$\sum_{m=1}^n \ln\left(1 - \frac{1}{m} + \frac{5}{4m ^ 2}\right) + \frac{1}{m} = \sum_{m = 1}^n \frac{1}{m^2}\left(\frac{1}{4} + \varepsilon_m\right)$$ The RHS converges, due to the convergence of $\sum_{m = 1}^\infty \frac{1}{m^2}$, to a limit we'll call $L$, and so by equality the $LHS$ converges to that same $L$.
Therefore, using the fact that we have a known expansion for $H_n := \sum_{m = 1}^n \frac{1}{m}$ : $$\begin{split}\sum_{m=1}^n \ln\left(1 - \frac{1}{m} + \frac{5}{4m ^ 2}\right) &= - H_n + L + o(1)\\ &= - \ln n - \gamma + L + o(1)\end{split}$$ where $\gamma$ is the Euler-Mascheroni constant.
This finally gives: $$\ln\left(n\prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)\right) = \ln n + \sum_{m=1}^n \ln\left(1 - \frac{1}{m} + \frac{5}{4m ^ 2}\right) = L - \gamma + o(1)$$ Consequently, when taking the $\exp$: $$n\prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right) \xrightarrow[n \to \infty]{} e^{L - \gamma}$$

The problem with my approach however is that I was not able to say anything about that limit $L$, but maybe someone will be able to evaluate that expression.

Answer 3

First let's express the partial products in terms of the Beta Function $$n\prod_{m=1}^n\left(1-\frac{1}{m}+\frac{5}{4m^2}\right) = n\prod_{m=1}^n\left(\frac{(2m-1)^2+4}{4m^2}\right) = n\prod_{m=1}^n\left(\frac{(m-1/2)^2+1}{m^2}\right)$$ $$=n\prod_{m=1}^n\left(\left(\frac{m-1/2-i}{m}\right)\left(\frac{m-1/2+i}{m}\right)\right)=n\frac{\left(n-\frac{1}{2}-i\right)!\left(n-\frac{1}{2}+i\right)!}{n!^2\left(-\frac{1}{2}-i\right)!\left(-\frac{1}{2}+i\right)!}$$ $$=n\frac{\Gamma\left(n+\frac{1}{2}-i\right)\Gamma\left(n+\frac{1}{2}+i\right)}{\Gamma(n+1)^2\Gamma\left(\frac{1}{2}-i\right)\Gamma\left(\frac{1}{2}+i\right)}=\frac{1}{n}\frac{\Gamma\left(n+\frac{1}{2}-i\right)\Gamma\left(n+\frac{1}{2}+i\right)}{\Gamma(n)^2\Gamma\left(\frac{1}{2}-i\right)\Gamma\left(\frac{1}{2}+i\right)}$$ $$=\frac{1}{n B\left(n,\frac{1}{2}-i\right)B\left(n,\frac{1}{2}+i\right)} \text{ where } B(\alpha,\beta) \text{ is the Beta function}$$ Now notice that, by Stirling's approximation, we have that $B(n,z_0)\sim\Gamma(z_0)n^{-z_0}$ so $$\sqrt{n}B\left(n,\frac{1}{2}+i\right)\sqrt{n}B\left(n,\frac{1}{2}-i\right)\sim n\Gamma\left(\frac{1}{2}+i\right)\Gamma\left(\frac{1}{2}-i\right)n^{-1/2-i}n^{-1/2+i}$$ $$=\Gamma\left(\frac{1}{2}+i\right)\Gamma\left(\frac{1}{2}-i\right)=\Gamma\left(\frac{1}{2}+i\right)\Gamma\left(1-\left(\frac{1}{2}+i\right)\right)=\frac{\pi}{\sin\left(\pi\left(\frac{1}{2}+i\right)\right)}$$ $$=\frac{\pi}{\cos(i\pi)}=\frac{\pi}{\cosh(\pi)}$$ which let's us compute the limit of the denominator and, thus, $$\lim_{n\to\infty}n\prod_{m=1}^n\left(1-\frac{1}{m}+\frac{5}{4m^2}\right) = \frac{\cosh(\pi)}{\pi}=\frac{e^\pi+e^{-\pi}}{2\pi}\quad \square$$

Answer 4

Preliminaries

The recursive relation for Gamma, $x\Gamma(x)=\Gamma(x+1)$, leads to $$ \prod_{m=1}^n\frac{m+x}m=\frac{\Gamma(n+1+x)}{n!\Gamma(1+x)}\tag1 $$ A limit that follows from Gautschi's Inequality $$ \lim_{n\to\infty}\frac{\Gamma(n+1+x)}{n!\,(n+1)^x}=1\tag2 $$ Thus, we can derive $$ \begin{align} &\prod_{m=1}^\infty\left(1+\frac xm\right)\left(1+\frac ym\right)\left(1-\frac {x+y}m\right)\\ &=\lim_{n\to\infty}\frac{\Gamma(n+1+x)}{n!\Gamma(1+x)}\frac{\Gamma(n+1+y)}{n!\Gamma(1+y)}\frac{\Gamma(n+1-x-y)}{n!\Gamma(1-x-y)}\tag{3a}\\ &=\lim_{n\to\infty}\frac{(n+1)^x}{\Gamma(1+x)}\frac{(n+1)^y}{\Gamma(1+y)}\frac{(n+1)^{-x-y}}{\Gamma(1-x-y)}\tag{3b}\\ &=\frac1{\Gamma(1+x)\Gamma(1+y)\Gamma(1-x-y)}\tag{3c} \end{align} $$ Explanation:
$\text{(3a):}$ apply $(1)$
$\text{(3b):}$ apply $(2)$
$\text{(3c):}$ take the limit

We will also use Euler's Reflection Formula $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag4 $$

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Computing the Product $$ \begin{align} \lim_{n\to\infty}n\prod_{m=1}^n\left(1-\frac1m+\frac5{4m^2}\right) &=\lim_{n\to\infty}\frac{n}{n+1}\prod_{m=1}^n\frac{m^2-m+\frac54}{m^2}\frac{m+1}m\tag{5a}\\ &=\prod_{m=1}^\infty\frac{\left(m^2-m+\frac54\right)(m+1)}{m^3}\tag{5b}\\ &=\prod_{m=1}^\infty\left(1-\frac{\frac12+i}m\right)\left(1-\frac{\frac12-i}m\right)\left(1+\frac1m\right)\tag{5c}\\[3pt] &=\frac1{\Gamma\!\left(\frac12-i\right)\Gamma\!\left(\frac12+i\right)\Gamma(2)}\tag{5d}\\ &=\frac{\sin\left(\frac\pi2-i\pi\right)}\pi\tag{5e}\\[6pt] &=\bbox[5px,border:2px solid #C0A000]{\frac{\cosh(\pi)}\pi}\tag{5f} \end{align} $$ Explanation:
$\text{(5a):}$ $\prod\limits_{m=1}^n\frac{m+1}m=n+1$
$\text{(5b):}$ $\lim\limits_{n\to\infty}\frac{n}{n+1}=1$
$\text{(5c):}$ $m^2-m+\frac54=\left(m-\frac12-i\right)\left(m-\frac12+i\right)$
$\text{(5d):}$ apply $(3)$
$\text{(5e):}$ apply $(4)$
$\text{(5f):}$ $\sin\left(\frac\pi2-ix\right)=\cos(ix)=\cosh(x)$

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Asymptotic Approximation

Using power series and the Euler-Maclaurin Sum Formula, we can extend Claude Leibovici's asymptotic expansion. $$ \begin{align} n\prod_{m=1}^n\left(1-\frac1m+\frac5{4m^2}\right) &=\frac{n}{n+1}\prod_{m=1}^n\frac{m^2-m+\frac54}{m^2}\frac{m+1}m\tag{6a}\\ &=\left(1+\frac1n\right)^{-1}\prod_{m=1}^n\left(1+\frac1{4m^2}+\frac5{4m^3}\right)\tag{6b} \end{align} $$ Explanation:
$\text{(6a):}$ $\prod\limits_{m=1}^n\frac{m+1}m=n+1$
$\text{(6b):}$ simplify

Using the power series for $\log(1+x)$ gives $$ \begin{align} &\log\left(1+\frac1{4m^2}+\frac5{4m^3}\right)\\ &=\frac1{4m^2}+\frac5{4m^3}-\frac1{32m^4}-\frac5{16m^5}-\frac{149}{192m^6}+O\left(\frac1{m^7}\right)\tag7 \end{align} $$ Applying the Euler-Maclaurin Sum Formula to $(7)$ yields $$ \begin{align} &\log\left(\prod_{m=1}^n\left(1+\frac1{4m^2}+\frac5{4m^3}\right)\right)\\ &=\log\left(\frac{\cosh(\pi)}\pi\right)-\frac1{4n}-\frac1{2n^2}+\frac{19}{32n^3}-\frac1{4n^4}+\frac{17}{960n^5}+O\left(\frac1{n^6}\right)\tag8 \end{align} $$ The Euler-Maclaurin Sum Formula adds a constant, which can be determined from $(5)$.

Subtracting the power series for $\log\left(1+\frac1n\right)$ from $(8)$ gives $$ \begin{align} &\log\left(\frac{n}{n+1}\prod_{m=1}^n\left(1+\frac1{4m^2}+\frac5{4m^3}\right)\right)\\ &=\log\left(\frac{\cosh(\pi)}\pi\right)-\frac5{4n}+\frac{25}{96n^3}-\frac{35}{192n^5}+O\left(\frac1{n^6}\right)\tag9 \end{align} $$ Using $(6)$, $(9)$, and the power series for $e^x$, we get $$ \begin{align} &n\prod_{m=1}^n\left(1-\frac1m+\frac5{4m^2}\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\cosh(\pi)}\pi\left(1-\frac5{4n}+\frac{25}{32n^2}-\frac{25}{384n^3}-\frac{1375}{6144n^4}-\frac{35}{8192n^5}+O\left(\frac1{n^6}\right)\right)}\tag{10} \end{align} $$ The error at $n=3$ is $0.036\%$; at $n=4$, it is $0.006\%$.

Answer 5

If we complex factor your product;

$$\dfrac{(2m - 1)^2 + 4}{4m^2} = \dfrac{(2m - 1)^2 - (2i)^2}{4m^2} = \dfrac{2m - 1 + 2i}{2m} \cdot \dfrac{2m - 1 - 2i}{2m}$$.

Hence, we split into $2$ products:

$$\prod_{m = 1}^{n}{1 - \dfrac{1}{m} + \dfrac{5}{4m^2}} = \prod_{m = 1}^{n}{\dfrac{2m - 1 + 2i}{2m} \cdot \dfrac{2m - 1 - 2i}{2m}} \\= \prod_{m = 1}^{n}{\dfrac{2m - 1 + 2i}{2m}} \cdot \prod_{m = 1}^{n}{\dfrac{2m - 1 - 2i}{2m}}$$

$\displaystyle\prod_{m = 1}^{n}{\dfrac{2m - 1 + k}{2m}} = \dfrac{\left(\dfrac{k}{2} + n - \dfrac{1}{2}\right)!}{\left(\dfrac{k}{2} - \dfrac{1}{2}\right)! \cdot n!}$

and

$\displaystyle\prod_{m = 1}^{n}{\dfrac{2m - 1 - k}{2m}} = \dfrac{\left(\dfrac{-k}{2} + n - \dfrac{1}{2}\right)!}{\left(\dfrac{-k}{2} - \dfrac{1}{2}\right)! \cdot n!}$

Hence,

$$n\cdot\prod_{m = 1}^{n}{1 - \dfrac{1}{m} + \dfrac{5}{4m^2}} = n\cdot\dfrac{\left(\dfrac{k}{2} + n - \dfrac{1}{2}\right)!}{\left(\dfrac{k}{2} - \dfrac{1}{2}\right)! \cdot n!}\cdot\dfrac{\left(\dfrac{-k}{2} + n - \dfrac{1}{2}\right)!}{\left(\dfrac{-k}{2} - \dfrac{1}{2}\right)! \cdot n!}$$ for which the limit approaches $(n \to \infty)$ $$\dfrac{1}{\left(\dfrac{-k}{2} - \dfrac{1}{2}\right)! \cdot \left(\dfrac{k}{2} - \dfrac{1}{2}\right)!}$$

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Substitute $k = 2i$; $$\dfrac{1}{\left(\dfrac{-2i}{2} - \dfrac{1}{2}\right)! \cdot \left(\dfrac{2i}{2} - \dfrac{1}{2}\right)!} \\= \dfrac{1}{\Gamma{\left(-i + \dfrac{1}{2}\right)} \cdot \Gamma{\left(i + \dfrac{1}{2}\right)}}$$

Edit;

$$\\= \dfrac{1}{\Gamma{\left(1 - \left(i + \dfrac{1}{2}\right)\right)} \cdot \Gamma{\left(i + \dfrac{1}{2}\right)}} \\= \dfrac{\sin{\left(\pi\left(i + \dfrac{1}{2}\right)\right)}}{\pi} = \dfrac{\cos{i\pi}}{\pi} = \dfrac{\cosh{\pi}}{\pi} \\\quad\\ \\\approx 3.6898$$

Answer 6

$$1 - \frac1m + \frac5{4m ^ 2}=\frac {m^2-m+\frac 54}{m^2}=\frac {(m-a)(m-b)}{m^2}$$ where $$a=\frac 12-i \qquad \text{and} \qquad b=\frac 12+i$$

Using Pochhammer symbols and their equivalents to the gamma function

$$P_n=\prod_{m=1}^n \left(1 - \frac1m + \frac5{4m ^ 2} \right)=\frac{\Gamma (n+1-a) \,\Gamma (n+1-b)}{\Gamma (1-a)\, \Gamma (1-b)\, \Gamma (n+1)^2}$$ Replacing partly $a$ and $b$ by their values $$P_n=\frac{\cosh (\pi )}{\pi }\,\frac{\Gamma (n+a) \,\Gamma (n+b)}{ \Gamma (n+1)^2}$$ Considering now $$Q_n=\log\left(n\,\frac{\pi }{\cosh (\pi )}\,P_n\right)$$ and using Stirling approximation $$Q_n=-\frac{5}{4 n}\left(1-\frac{5}{24 n^2}+\frac{7}{48 n^4}+O\left(\frac{1}{n^6}\right)\right)$$ Back to $P_n$ $$n\,P_n=\frac{\cosh (\pi )}{\pi }\,\left(1-\frac{5}{4 n}+\frac{25}{32 n^2}-\frac{25}{384 n^3}+O\left(\frac{1}{n^4}\right)\right)$$ which is a good approximation even for small values of $n$. For example, the absolute relative error is $0.045$% for $n=5$.

Answer 7

We have that

$$n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)=e^{\log n +\sum_{m = 1} ^ n \log\left(1 - \frac1m + \frac5{4m ^ 2}\right)}$$

and

$$\sum_{m = 1} ^ n \log\left(1 - \frac1m + \frac5{4m ^ 2}\right)=\log \frac 5 4+\sum_{m = 2} ^ n \log\left(1 - \frac1m \right)+\sum_{m = 2} ^ n \log\left(1 + \frac5{4m ^ 2-4m}\right)$$

with

$$\sum_{m = 2} ^ n \log\left(1 - \frac1m \right)=-\log n $$

and numerically

$$\sum_{m = 2} ^ n \log\left(1 + \frac5{4m ^ 2-4m}\right)\to s\approx 1.0824\tag 1$$

therefore

$$n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)\to e^{s+\log \frac 54}=\frac 54 e^{s}\approx 3.68$$

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Finally, using the fine result given by jjagmath, now we are able to find a closed form for $(1)$, indeed

$$\prod_{m = 2} ^ n \left(1 + \frac5{4m ^ 2-4m}\right)=\prod_{m = 2} ^ n \frac{4m ^ 2-4m+5}{4m ^ 2-4m}=\prod_{m = 2} ^ n \frac{(2m-1)^2+4}{(2m-1)^2}\frac{(2m-1)^2}{2m(2m-2)}=$$

$$=\prod_{m = 2} ^ n \left(1+\frac{4}{(2m-1)^2}\right)\frac2{\displaystyle \prod_{m = 2} ^ n\frac{4(m-1)^2}{4(m-1)^2-1}} \to e^s=\frac15\cosh(\pi) \frac 4 \pi =\frac 4 5 \frac{\cosh(\pi)}{\pi}$$

which leads to the limit $\frac{\cosh(\pi)}{\pi}$.

Answer 8

$$L=\lim_{n \to \infty} n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)$$

Let $\color{blue}{t = -\frac{1}{2} + i}$ and $\color{blue}{\bar{t} = -\frac{1}{2} - i}$

Consider this expression,

$$e^{\gamma\cdot t}\cdot e^{\gamma\cdot \bar{t}}\cdot e^{-\frac{t}{m}}\cdot e^{-\frac{\bar{t}}{m}} \prod_{m=1}^n \left(1 + \frac{t}{m}\right) \prod_{m=1}^n \left(1 + \frac{\bar{t}}{m}\right)$$

Where $\gamma$ is the Euler Mascheroni constant

$$ = e^{H_n-\gamma} \prod_{m=1}^n \left|1 + \frac{t}{m} \right|^2$$

$$ = e^{H_n-\gamma} \prod_{m=1}^n \left(1 - \frac{1}{m} + \frac{5}{4m^2}\right)$$

$$ \underset{\times \frac{n}{n}}= \frac{e^{H_n-\gamma}}{n} n \prod_{m=1}^n \left(1 - \frac{1}{m} + \frac{5}{4m^2}\right)$$

$$\color{red}{H_n \approx \log n + \gamma\implies \frac{e^{H_n-\gamma}}{n}\to 1}$$ From here,

\begin{equation*} \boxed{\frac{1}{\Gamma(z)} = ze^{\gamma z} \prod_{m=1}^{\infty} \left( 1 + \frac{z}{m} \right) e^{-z/m}\implies \frac{1}{z\Gamma(z)} = e^{\gamma z} \prod_{m=1}^{\infty} \left( 1 + \frac{z}{m} \right) e^{-z/m}} \end{equation*}

Hence,

\begin{align*} L &= \left(\frac{1}{t\Gamma(t)}\right)\left(\frac{1}{\bar{t}\Gamma(\bar{t})}\right) \\ &= \left(\frac{1}{\Gamma(1+t)}\right)\left(\frac{1}{\Gamma(1+\bar{t})}\right) \end{align*}

• From here,

• $\Gamma(n+1)=n\Gamma(n)$

Using this,

• $1+t=\frac{1}{2}+i$

• $1+\bar{t}=\frac{1}{2}-i$

• $\cos(ix)=\cosh(x)$

\begin{align*} L &= \frac{\sin(\pi(1+t))}{\pi} \\ L &= \frac{\cos(i\pi)}{\pi} \end{align*}

$${\color{red}{\therefore\lim_{n\to\infty} n \prod_{m=1}^n \left( 1 - \frac{1}{m} + \frac{5}{4m^2} \right)=\frac{\cosh(\pi)}{\pi}}}$$