Hey persons, I was wondering if it is possible to solve for k

>>8581136
[math]\displaystyle \int_k^{\infty} \frac{1}{\Gamma(x + 1)} \; dx = e[/math] is what you want, unless you're serious about having a domain [math]\displaystyle (k, \infty) \cap \mathbb{N}[/math].

>>8581152
Yes, I was wondering about naturals, probably only because of my lack of understanding.
x! for x > -1 is well-defined though, correct?
I guess this is beyond me, but now I have reading material on the gamma function.
Thank you anyway :)

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Using the Gamma function
k = -0.597
is quite close

>>8581189
Yea that's how far I got with integrating up to 10^6 of x! and playing around with manipulate in Mathematica for the lower bound.
Crude, but effective, I guess.

>>8581189
>597
that looks very familiar
more digits?

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>>8581206
>>8581207
forgot the picture.

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There's a lot of k's that work OP.

>>8581207
>that looks very familiar
>more digits?
Just type it into Wolfram Alpha and it will give you some closed form approximations.

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>>8581206
I take it you started with
[math] \sum_{x=0}^\infty \dfrac{1}{x!} = e [/math]
and passed over to
[math] \int_{x=0}^\infty \dfrac{1}{x!} [/math]

That's the right time for Ramanuja magic, telling you

[math] \int_{x=0}^\infty \dfrac{1}{x!} = e - \left( \dfrac{1}{2} - \dfrac{1}{12} \dfrac{d}{dx} \dfrac{1}{x!} + ... \right) |_{x=0} [/math]

where the dots are higher order derivatives. For the Gamma function as x!, the derivative is the Euler–Mascheroni constant and this first order approximation misses the numerical integral by 0.00015.

>>8581207
-0.597006

>>8581233
ad: A representation of [math] \dfrac {1} {x!} [/math] that's equivalent to that of the Gamma function but possibly more approachable is

[math] \dfrac {1} {x!} = e^{x \gamma} \prod_{n=1}^\infty \left( 1 + \dfrac {x} {n} \right) e^{-x/n} [/math]

(where [math] \gamma [/math] is again the Euler–Mascheroni constant)

>>8581233
Could you explain how to apply "Ramanujan magic" to arrive at that result?

>>8581136
Alright OP I suggest you start learning about measure and integration theory. It sounds like something you'd enjoy. Using measure theory and the Lebesgue integral, we can quite easily find that when using the counting measure, k = 0.

>>8581189
>>8581216
>>8581233

Thank you.
Do you know of a way to get to actually solving for k, or am I stuck with approximations forever?

>>8581233

I'd love it if you explained yourself further.