How to prove $\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$

How can I prove that

$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$

The result is given here (result 48).

The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.

Some of my attempts include:

1. Multiplying known infinite products for $$\frac{\pi}{2}$$ and $$\frac{1}{e}$$

$$\frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots}$$ $$\frac{1}{e} = \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$

as well as others given in https://arxiv.org/pdf/1005.2712.pdf

1. Trying to find the partial products from $${n=1}$$ to $$k$$, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}}$$

However, I have been unsuccessful producing partial products on my own.

1. Taking the $$\ln$$ of the product to try to find the partial sums;

2. Trying to transform the following integral into infinite product.

$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e}$$

From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.

Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.

Thank you kindly for your help and time.

Melzak (1961) shows that the largest-volume cylinder (Cartesian product of a hypersphere and a line) in an $$n$$-dimensional unit sphere occupies this proportion of the sphere: $$\rho_n=\frac{2(n\pi)^{-1/2}(1-1/n)^{(n-1)/2}\Gamma(n/2+1)}{\Gamma((n+1)/2)}$$ We have $$\rho_2=\frac2\pi$$ and $$\lim_{n\to\infty}\rho_n=\sqrt{\frac2{\pi e}}$$. Now define $$\sigma_n=\frac{\rho_{n+2}}{\rho_n}=\sqrt{\left(\frac n{n+2}\right)^n\left(\frac{n+1}{n-1}\right)^{n-1}}$$ Telescoping on $$\sigma_n$$ for $$n=2,4,6\dots$$ then gives $$\sqrt{\frac\pi{2e}}=\prod_{n=1}^\infty\left(\frac n{n+1}\right)^n\left(\frac{2n+1}{2n-1}\right)^{(2n-1)/2}$$ Squaring gives $$\frac\pi{2e}=\prod_{n=1}^\infty\left(\frac {2n}{2n+2}\right)^{2n}\left(\frac{2n+1}{2n-1}\right)^{2n-1}=\prod_{n=1}^\infty(1+2/n)^{(-1)^{n+1}n}$$

• Why is this product convergent? We don't have $\lim_{n\to\infty}(1+2/n)^{(-1)^{n+1}n}=1$ Commented Jan 7, 2021 at 17:06
• @bjorn93 Pairs of terms multiply to one. Commented Jan 8, 2021 at 3:19
• Then the product should be defined as $\prod_{n=1}^{\infty} a_{2n-1}a_{2n}$ where $a_n$ is the current term. I noticed that this is what you actually have in the last line. The way it's defined now means the product is divergent. Commented Jan 8, 2021 at 4:05

This infinite product could be computed (more or less) directly.

At first, let's compute infinite sum instead of infinite product: $$P=\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}\,\,n}$$ $$S = \ln P = \sum_{n=1}^\infty (-1)^{n+1}n \ln\left(1+\frac2n\right) = \sum_{n=1}^\infty (-1)^{n+1}a_n$$ At first we need some regularization: $$a_n=2-\frac2n+\frac{8}{3n^2}+\mathcal O(\frac{1}{n^3}) \nrightarrow 0$$ Nevertheless we could only consider partial sums of the $$S$$ with odd (or even) $$n$$; in other words, we'll assume that $$\sum_{n=1}^\infty (-1)^{n+1} = 0$$. (I know, I know, this is a not rigorous, but let me advance).

Ok, let's convert $$a_n$$ to power series: $$a_n = \sum_{k=0}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}=2+\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k},$$ so we need to compute $$S = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}$$ (look, we «cancel» 2 in $$a_n$$'s expansion). Let's change summation order: $$S = \sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1} \left(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k}\right).$$ Series in the parentheses is alternating zeta-function: $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k} = \left(1-\frac{1}{2^{k-1}}\right)\zeta(k),$$ where $$\zeta(k)$$ is the Riemann zeta function ($$k>1$$). At $$k=1$$ we have $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\ln2.$$

So, $$S = -2\ln2 + \sum_{k=2}^\infty (-1)^k\frac{2^{k+1}}{k+1}\left(1-\frac{1}{2^{k-1}}\right)\zeta(k)\\=-2\ln2+\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\zeta(k).$$ This series seems hopeless, but zeta function involves in many other series. We'll use this: $$\Gamma(s)\zeta(s)=\int\limits_0^\infty \!\frac{x^{s-1}}{e^x-1}dx.$$

Hence $$S + 2\ln2 = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{\Gamma(k)}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx \\= \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{(k-1)!}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx,$$ or $$T = S + 2\ln2 = \int\limits_0^\infty \!\frac{dx}{e^x-1} \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!}$$

The last series could be computed by standard methods via series for exponent and differentiation w.r.t. $$x$$: $$\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!} = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{(k+1)!}kx^{k-1}\\= \frac{3}{x^2}+\frac{1}{x^2}e^{-2x}+\frac{2}{x}e^{-2x}-\frac{4}{x^2}e^{-x}-\frac{4}{x}e^{-x}$$

Let's collect all these pieces together: $$T = \int\limits_0^\infty \frac{e^{-2x}dx}{x^2(e^x-1)}(3e^{2x}-4(1+x)e^x+2x+1).$$ We have here terms with powers of $$\mathrm{exp}$$ and $$x$$. For me simplest and straightforward method for calculation of this integral is expansion of denominator: $$\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=e^{-x}\sum_{m=0}^\infty e^{-mx} = \sum_{m=1}^\infty e^{-mx}.$$ So we have $$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-2x}\sum_{m=1}^\infty e^{-mx},$$ or $$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} \sum_{m=3}^\infty e^{-mx}.$$ Let's change summation and integration: $$T = \sum_{m=3}^\infty \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$

Now we need to compute $$I_m = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$ Maple says $$I_m = m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$ We could believe in it or could to get it:) At first, $$\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} = \sum_{k=0}^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k,$$ and $$\int\limits_0^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k e^{-mx} dx = \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$ (we used Euler's formula for $$\Gamma(x)$$). So, $$I_m = \sum_{k=1}^\infty \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$ (term at $$k=0$$ vanishes). Now we can diff. w.r.t. $$m$$: $$\frac{d^2 I_m}{dm^2}= \sum_{k=1}^\infty (12\cdot 2^k-4k-12) \frac{1}{m^{k+3}} = \frac{1}{m^3} \sum_{k=1}^\infty \frac{12\cdot 2^k}{m^k}-\frac{4k}{m^k}-\frac{12}{m^k},$$ or $$\frac{d^2 I_m}{dm^2}= \frac{1}{m^3} \left(\frac{24}{m - 2} - \frac{4m}{(m - 1)^2} - \frac{12}{m - 1}\right)$$ (while $$m>2$$). Integrate twice and we have indeed $$I_m = (3m - 6)\ln(m - 2) + (-4m + 8)\ln(m - 1) + 2 + (m - 2)\ln(m)\\= m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$ We start summation from $$m=3$$; let $$m=2+p$$: $$I_{p+2} = p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2$$

Now we have $$T = \sum_{p=1}^\infty p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2 = \sum_{p=1}^\infty t_p.$$ We could find approximation for this sum by some methods, but I prefer more detailed calculation. At first, let's rearrange $$t_p$$: $$t_p = [(p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p)] \\ +[2-2\ln(p + 2) + 4\ln(p + 1) ].$$ First sum could be found by kind of telescoping summation: $$T^1_N = \sum_{p=1}^N (p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p) \\= -3(N + 1)\ln(N + 1) + (N + 2)\ln(N + 2) - 2\ln(2)$$ (put $$f(x)=x\ln x$$ and look at partial sums). Analogously for second term we have $$T^2_N = \sum_{p=1}^N 2-2\ln(p + 2) + 4\ln(p + 1) \\= 2\sum_{p=1}^N \ln p + 2\ln(N + 1) - 2\ln(N + 2) + 2\ln(2) + 2N$$

Finally we have $$T_N = 2\sum_{p=1}^N \ln p - (3N + 1)\ln(N + 1) + N\ln(N + 2) + 2N.$$ Asymptotic expansion for the first series is well-known and could be found from Euler—Maclaurin: $$\sum_{p=1}^N \ln p = N(\ln(N) - 1) + \frac12\ln(2\pi) + \frac12\ln(N) + \frac{1}{12N}+\mathcal O \left(\frac{1}{N^2}\right),$$ and finally (really finally :) $$T_N = \ln(2\pi)-1 - \frac{4}{3N}+\mathcal O \left(\frac{1}{N^2}\right).$$

So, $$T = \ln(2\pi)-1,$$ $$S = T - 2\ln2 = \ln\frac{\pi}{2}-1 = \ln\frac{\pi}{2e},$$ and $$P = \frac{\pi}{2e}.$$

• Wowzers , this is going to take a while to digest. I really appreciate the time you took to give this answer. Cheers! Commented Jan 8, 2021 at 4:22
• @No-oneImportant calculations above need some clarification and should be more rigorous (we always change summation and integration order, for example), and they are quite boring, but all they are very straightforward :) Commented Jan 8, 2021 at 12:53
• The zeta series you called hopeless occurs in the log of Stirlings approx. polynomal, and is quickly derived ^^ saves a whole lot of work. But hats off for solving it regardless! Commented Feb 7, 2021 at 4:30

The stated product does not converge and therefor the answer is incorrect.

It's only valid if the final value is even.

$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \neq \frac{\pi}{2e}$$

Most likely you ment:

$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} = \frac{\pi}{2e}$$

For the proof of even value's or the product above: Consider: $$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} =$$

$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} \big(n \ln(1+2/n)-2\big)\bigg)$$ We know the factorial (e.g. stirlings formula, but people don't seem to write it this way, derived by writing it as sums, alternatively you can use divergent products, see: Divergent products.)

$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2 n\pi}\big) \exp\bigg(-\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg)$$

$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} 2\ln\bigg(\frac{(1+2/n)^{n/2})}{e}\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2-1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n-1/2})}{e}\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n})}{e}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1)^j}-\frac{\zeta(-j)}{j(n)^j}\bigg)-\bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1/2)^j}-\frac{\zeta(-j)}{j(n-1/2)^j}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$

if $$\Delta_n^+=f(n)-f(n+1)$$ and if f(n)= 0 as n goes to infty.

$$\exp\bigg(2\sum_{n=1}^{\infty} \sum_{j=1}^\infty \Delta_n^+\frac{\zeta(-j)}{j(n)^j}-\Delta_n^+\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n-1/2)^j}- \Delta_n^+\frac{1}{2}\ln\big(\frac{n-1/2}{n}\big)\bigg)=$$

$$\exp 2 \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j}-\sum_{j=1}^\infty \frac{\zeta(-j)}{j(1/2)^j}+\frac{\ln(2)}{2}\bigg)=$$ $$\exp 2 \bigg(\ln\big(\frac{\sqrt{2\pi}}{e}\big)-\ln(\frac{\sqrt{e}}{2\sqrt{2}})+\frac{\ln(2)}{2}\bigg)=$$

$$\frac{2\pi}{e^2}\frac{e}{8} 2=\frac{\pi}{2e}$$

Maybe try splitting it into two products to get rid of the alternating sign: $$\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}n}=\prod \left(1+\frac2{2n}\right)^{-2n}\prod\left(1+\frac2{2n+1}\right)^{2n+1}$$ then look at these individually, the first one: $$P_1=\prod_{n=1}^\infty\left(1+\frac1n\right)^{-2n}=\left[\prod_{n=1}^\infty\left(1+\frac1n\right)^n\right]^{-2}$$ although this diverges you may be able to get it to cancel with something from the other expression?

So far these are just some thoughts.

Introduction:

We can use the product $$\sin\pi x=\pi x\prod_{n\ge1}(1-x^2/n^2)$$ to show that $$\mathrm L(x)=\int_0^x\ln\sin\pi t\,dt=\ln\left[\left(\frac{\pi x}{e}\right)^x\prod_{n\ge1}e_k(x)\right],\tag1$$ where $$e_k(x)=\frac{j(n+x)}{(en)^{2x}j(n-x)},$$ and $$j(x)=x^x$$. This is done by expanding $$\ln\sin\pi t=\ln(\pi t)+\sum_{n\ge1}\ln(1-t^2/n^2)$$ and integrating termwise. On the other hand, we know a lot about a closely related function called the Clausen function, which is defined as $$\mathrm{Cl}_2(x)=-\int_0^x\ln\left|2\sin\tfrac{t}{2}\right|dt=\sum_{n\ge1}\frac{\sin nx}{n^2}.\tag2$$ We can show that $$\tau(x)=\prod_{n\ge1}e_n(x)=\left(\frac{e}{\pi x}\right)^x\exp\mathrm L(x)=\left(\frac{e}{2\pi x}\right)^x\exp\left[-\frac{1}{2\pi}\mathrm{Cl}_2(2\pi x)\right].\tag3$$

Main Ideas: Consider the value of $$\tau(1)$$. We have $$\mathrm{Cl}_2(2\pi)=\sum_{n\ge1}\frac{\sin2\pi n}{n^2}=0,$$ so $$\tau(1)=\frac{e}{2\pi}.$$ We will consider the convergent analog of the divergent product in the title $$P=\prod_{n\ge1}\left(\frac{n}{n+1}\right)^{2n}\left(\frac{2n+1}{2n-1}\right)^{2n-1}.$$ We aim to show that $$P=\frac{1}{4\tau(1)}=\frac{\pi}{2e}$$.

From the Clausen duplication formula and a bit of algebraic manipulation, we may show that $$\tau(2x)=\frac12\left(\frac{\pi}{e}\right)^{2x-1}\left(\frac{\tau(x)}{j(\tfrac12-x)\tau(\tfrac12-x)}\right)^2.\tag4$$ If we can find a suitable value $$x$$ to plug into $$(4)$$, we may be able to get an expression for $$\frac{1}{4\tau(1)}$$ which can be transformed, via algebraic manipulations, to the expression for $$P$$. I will update my answer once a suitable value is found.