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I can show the convergence of the following infinite product and some bounds for it:

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots<$$

$$<\left(1+\frac{1}{4} \right)\left(1+\frac{1}{9} \right)\left(1+\frac{1}{16} \right)\cdots=\prod_{k \geq 2} \left(1+\frac{1}{k^2} \right)=\frac{\sinh \pi}{2 \pi}=1.83804$$

Here I used Euler's product for $\frac{\sin x}{x}$.

The next upper bound is not as easy to evaluate, but still possible, taking two more terms in Taylor's series for $\sqrt[k]{1+\frac{1}{k} }$:

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{k-1}{2k^4}+\frac{2k^2-3k+1}{6k^6} \right)=$$

$$=\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{1}{2k^3}+\frac{5}{6k^4}-\frac{1}{2k^5}+\frac{1}{6k^6} \right)<$$

$$<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{108}+\frac{\pi^6}{5670}-1-\frac{\zeta (3)}{2}-\frac{\zeta (5)}{2} \right)=1.81654$$

The numerical value of the infinite product is approximately:

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=1.758743628$$

The ISC found no closed from for this number.

Is there some way to evaluate this product or find better bounds in closed form?


Edit

Clement C suggested taking logarithm and it was a very useful suggestion, since I get the series:

$$\ln \prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}= \frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots$$

I don't know how to find the closed form, but I can certainly use it to find the boundaries (since the series for logarithm are very simple).

$$\frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots>\sum^{\infty}_{k=2} \frac{1}{k^2}-\frac{1}{2}\sum^{\infty}_{k=2} \frac{1}{k^3}$$

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}-\frac{1}{2}-\frac{\zeta (3)}{2} \right)=1.72272$$

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{5}{6}-\frac{\zeta (3)}{2} \right)=1.77065$$

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{7}{12}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.75438$$

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{47}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.76048$$

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{37}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4} -\frac{\zeta (7)}{6}\right)=1.75803$$

This method generates much better bounds than my first idea. The last two are very good approximations.


Edit 2

Actually, would it be correct to write (it gives the correct value of the product):

$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left( \sum_{k \geq 2} \frac{(-1)^k \zeta(k)}{k-1} \right)$$

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  • $\begingroup$ My money is on $e^{2\pi/5}/2$. Taking the logarithm and considering instead a sum may help... and starting the product at $1$ instead of $2$ may also -- this is the factor $2$ in my guess.) $\endgroup$ – Clement C. Feb 22 '16 at 1:25
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    $\begingroup$ @ClementC., but it's different from the numerical value $\endgroup$ – Yuriy S Feb 22 '16 at 1:44
  • $\begingroup$ Euler Mac Laurin gives 1.73287 keeping only the first two terms $\endgroup$ – tired Feb 22 '16 at 2:16
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    $\begingroup$ The product evaluates to $\simeq 1.0011\cdot \frac{e^{2\pi/5}}{2}$ so the conjecture above seems to be ruled out (this was double checked by also evaluating the series representation). $\endgroup$ – Winther Feb 22 '16 at 2:29
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    $\begingroup$ I googled the sum (from 1) and found something. books.google.com/… i.imgur.com/w5WQE2E.png $\endgroup$ – Steve Kass Feb 22 '16 at 2:34
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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 $$ \prod_{k=2}^\infty\left(1+\frac1k\right)^{1/k}=1.75874362795118482469989684966 $$

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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.

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

Convergence


We can also obtain the following interesting equality:

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

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  • $\begingroup$ Re: "This same series also appeared [...] This is the same numerical value as:": showing equality of the two (series, not deriving their value) is not too hard -- I assume you don't need it, but in case you do it's a matter of 3-4 lines. $\endgroup$ – Clement C. Feb 22 '16 at 4:18
  • $\begingroup$ Which one? The 0.788530566 one? $\endgroup$ – Clement C. Feb 22 '16 at 4:24
  • $\begingroup$ I have nothing really conclusive in making a useful connection between the values... what sort of relation are you hoping for? $\endgroup$ – Clement C. Feb 22 '16 at 4:35
  • $\begingroup$ I don't know. It came up in my search and S. Finch hinted at some kind of connection. $\endgroup$ – Yuriy S Feb 22 '16 at 4:37
  • $\begingroup$ $$\begin{align} \sum_{k=2}^\infty\frac{1}{k}\ln\left(1+\frac{1}{k}\right) &=\sum_{k=2}^\infty\frac{\ln(k+1)-\ln(k)}{k}\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N}\frac{\ln(k+1)}{k}-\sum_{k=2}^{N}\frac{\ln(k)}{k}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N}\frac{\ln(k+1)}{k}-\sum_{k=1}^{N-1}\frac{\ln(k+1)}{k+1}\right)\\ &=\lim_{N\to\infty}\left(\frac{\ln(N+1)}{N}+\sum_{k=2}^{N-1}\frac{\ln(k+1)}{k(k+1)}-\frac{\ln(2)}{2}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N-1}\frac{\ln(k+1)}{k(k+1)}\right)-\frac{\ln(2)}{2}\\ &=\sum_{k=1}^{\infty}\frac{\ln(k+1)}{k(k+1)}\\ \end{align}$$ $\endgroup$ – alex.jordan Feb 22 '16 at 5:15
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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.

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  • $\begingroup$ This seems to imply a relation to the derivative of the reimann zeta function: $\zeta ' (s) = -\sum_{n=1}^{\infty} \frac{\ln (n)}{n^s}$ $\endgroup$ – Jacob Apr 6 '16 at 2:32
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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.

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