Evaluate the infinite product $\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots$

Solution 1:

This is not an answer, but it's important and I post it separately from the question itself.

I found in this answer by @RandomVariable the following series:

$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1} $$

They are related to $\gamma_1$ - Stieltjes constant.

This same series also appeared in this paper by Steven Finch, page 5.

$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=1.2577468869$$

This is the same numerical value as:

$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=1.2577468869$$

Which is confirmed in this paper by the same author, page 3, where this form of the series is used.

It is connected to the integral (page 2, the same paper):

$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=-\int_{1}^{\infty} \frac{\ln (y-[y])}{y^2}dy$$

Where $[y]$ is the floor function, meaning $y-[y]$ is the fractional part of $y$.

In another paper this series is connected to the numer of divisors of $n!$, however slightly different integral representation is used (page 3):

$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\int_{1}^{\infty} \frac{\ln ([y]+1)}{y^2}dy$$

And finally, this is slightly related to Alladi-Grinstead Constant, which is given by:

$$e^{c-1}$$

$$c=\sum_{k=2}^{\infty} \frac{\ln (\frac{k}{k-1})}{k}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k+1}=0.788530566$$

See also the original Alladi and Grinstead paper here.

And this is also somehow connected to the Luroth series representations of real numbers.

Oh, and thanks to @SteveKass for this useful link.

Comparing the convergence of three series, we find that even though they are equivalent, the convergence rate is drastically different.

$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}$$

We can also obtain the following interesting equality:

$$(1+1)\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots=\sqrt{2} \sqrt[6]{3} \sqrt[12]{4} \sqrt[20]{5} \sqrt[30]{6} \cdots=\prod_{k=1}^{\infty}(k+1)^{\frac{1}{k(k+1)}}$$

Solution 2:

I don't know if a closed form exists, but to get geometric convergence, we can use the following.

Since the product starts at $k=2$, we compute the sum $$ \begin{align} \sum_{k=2}^\infty\frac1k\log\left(1+\frac1k\right) &=\sum_{k=2}^\infty\frac1k\sum_{n=1}^\infty\frac{(-1)^{n-1}}{nk^n}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\sum_{k=2}^\infty\frac1{k^{n+1}}\\ &=\sum_{n=1}^\infty(-1)^{n-1}\frac{\zeta(n+1)-1}{n}\\[6pt] &=0.564599706384424320592667709038 \end{align} $$ Note that $\frac{\zeta(n+1)-1}{n}\sim\frac1{n2^{n+1}}$. This gives better than geometric convergence.

Applying $e^x$, we get $$ \prod_{k=2}^\infty\left(1+\frac1k\right)^{1/k}=1.75874362795118482469989684966 $$

Solution 3:

As already discussed above by Yuriy S and others, the product is intimately linked with the series $\sum_{n=1}^{\infty} { \ln(n+1) \over n(n+1) } \approx 1.2577\dots$. The connection is derived as follows:

$$ \prod_{k=2}^{\infty} \left( 1+ {1\over k} \right)^{1/k} \\ =\exp \left( \ln \left( \prod_{k=2}^{\infty} \left( 1+ {1\over k} \right)^{1/k} \right) \right) \\ = \exp \left( \sum_{n=2}^{\infty} \ln \left( \left( 1+ {1\over k} \right)^{1/k} \right) \right) \\ = \exp \left( \sum_{n=2}^{\infty} { \ln\left({k+1 \over k}\right)\over k } \right) \\ =\exp \left( \sum_{n=2}^{\infty} {\ln(k+1)\over k } -\sum_{n=2}^{\infty} {\ln(k) \over k} \right)\\ =\exp \left( \sum_{n=3}^{\infty} {\ln(k) \over k-1} -\sum_{n=3}^{\infty} {\ln(k) \over k} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=3}^{\infty} {\ln(k) \over k(k-1)} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=2}^{\infty} {\ln(k+1) \over k(k+1)} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)} - {1 \over 2}\ln 2 - {1 \over 2}\ln 2 \right)\\ =e^{\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}} \cdot e^{-\ln 2}\\ ={1 \over 2}e^{\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}} $$

Therefore the product will only have a closed form if $\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}$ has a closed form.

Solution 4:

Working off of other people's findings, you can write $$\sum_{k=2}^{\infty} \frac{(-x)^k \zeta (k)}{k} = x \gamma + \ln (\Gamma(x+1))$$ $$\frac{d}{dx}\sum_{k=2}^{\infty} \frac{(-x)^k \zeta (k)}{k} = -\sum_{k=2}^{\infty} (-x)^{k-1} \zeta (k)=\gamma+\psi(x+1)=H_x$$ $$\sum_{k=2}^{\infty} (-x)^{k-2} \zeta (k)=\frac{H_x}{x}$$ $$\int_0^{x} \sum_{k=2}^{\infty} (-y)^{k-2}\zeta (k) dy = -\sum_{k=2}^{\infty} \frac{(-x)^{k-1}\zeta (k)}{k-1} = -\int_0^x \frac{H_y}{y} dy$$ $$\sum_{k=2}^{\infty} \frac{(-x)^{k}\zeta (k)}{k-1} = x\int_0^x \frac{H_y}{y} dy$$ However, I believe $\int_0^1 \frac{H_x}{x} dx$ has no closed form, meaning that $$\prod_{k=2}^{\infty} \sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left(\sum_{k=2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}\right)=\frac{1}{2}\exp \left({\int_0^1 \frac{H_x}{x} dx} \right)$$ has no closed form either.