Infinite summation for x^(1-x)

[Ashwin_Kumar]

Along with some friends, i was trying to calculate the end product of a series(x^(1-x)). I for a finite value, i got a number close to arcsinh(SQRT 6), and by comparing the functions i believe it converges at infinity. However, i am unable to 100% prove this, and i am asking if the summation is correct in particular and whether a similar series is already in existence.

..\infty
\sum\chi^{1-\chi}=arcsinh(\sqrt{6})
..\chi=1
(PS does anyone know how to type summations much faster than using the sigma notation, then typing below and above etc. I am not too familiar with Latex

Another summation that i would like to confirm:
..\infty
\sum\frac{1}{\Gamma\chi}=e
..\chi=1
This can be derived by representing e^x with a summation.

 

[dimension10]

Along with some friends, i was trying to calculate the end product of a series(x^(1-x)). I for a finite value, i got a number close to arcsinh(SQRT 6), and by comparing the functions i believe it converges at infinity. However, i am unable to 100% prove this, and i am asking if the summation is correct in particular and whether a similar series is already in existence.

..\infty
\sum\chi^{1-\chi}=arcsinh(\sqrt{6})
..\chi=1
(PS does anyone know how to type summations much faster than using the sigma notation, then typing below and above etc. I am not too familiar with Latex

Another summation that i would like to confirm:
..\infty
\sum\frac{1}{\Gamma\chi}=e
..\chi=1
This can be derived by representing e^x with a summation.

Oh, that thing we were discussing. I don't think so because the hyperbolic function at 1.6284 is zooming upwards, so it crosses 6.0002 rather than coming down to 6 itself.

The 2nd one is correct, you can verify it using one of Euler's stuff.

And there is a much simpler way than writing above sigma and below it. You can just use numberempire LaTeX. Of course, you can't copy paste the rendered equation. But just type the code here between [te.x] and [/te.x] (without dots)

For example, for typing \sum_{x=1}^{\infty}{x}^{1-x}=arcsinh\sqrt{6} type:

[tx] \sum_{x=1} ^ {\infty} {x} ^ {1-x} = arcsinh \sqrt {6} [/tx] (with e between t and x)

 

[Gib Z]

The equality does not hold. The summation is approximately 1.628473712901584447055889143261883031650540316 whilst archsinh sqrt6 is approximately 1.6283069774000262046581136802999892614654729597107490837603. I very much doubt the summation has a simple closed form.