Edit: In hindsight I should've posted this in analysis; I'm sorry I just didn't notice it there. Maybe a mod could move it there, thanks.

This proof starts out by considering the differential operator $\displaystyle D_t f(x) = \frac{d}{dt}\frac{d^t}{dx^t} f(x)$ which measures the growth of continued differentiation of f(x). It is spreadable across addition $\displaystyle D_t (f(x) + g(x)) = D_t f(x) + D_t g(x)$. And:

$\displaystyle D_t e^x = 0$; which is important for the proof

And next, using traditional fractional calculus laws:

$\displaystyle D_t x^n = \frac{d}{dt} \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t}$

which comes to, (if you want me to show you the long work out just ask, I'm trying to be brief)

$\displaystyle D_t x^n = x^{n-t}\frac{\Gamma(n+1)}{\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x))$ where $\displaystyle \psi_0(x)$ is the digamma function.

So now we do the fun part:

$\displaystyle D_t e^x = 0\\\\

D_t (\sum_{n=0}^{\infty} \frac{x^n}{n!}) = 0\\\\

\sum_{n=0}^{\infty} D_t \frac{x^n}{n!} = 0\\\\$

So we just plug in our formula for $\displaystyle D_t x^n$ and divide it by n!:

$\displaystyle 0 = \sum_{n=0}^{\infty} x^{n-t}\frac{\Gamma(n+1)}{n!\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x))$

(divide $\displaystyle \Gamma(n+1)$ by n!.)

we expand these and seperate and rearrange:

$\displaystyle 0 = \sum_{n=0}^{\infty} (\frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t) - \frac{x^{n-t}}{\Gamma(n+1-t)}\ln(x))\\\\

0 = (\sum_{n=0}^{\infty} \frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t)) - (\sum_{c=0}^{\infty} \frac{x^{c-t}}{\Gamma(c+1-t)}\ln(x))\\\\

\ln(x)(\sum_{c=0}^{\infty} \frac{x^{c-t}}{\Gamma(c+1-t)}) = (\sum_{n=0}^{\infty} \frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t))\\\\$

And now if you're confused what t represents, you'll be happy to hear we eliminate it now by setting it to equal 0. therefore, all our gammas are factorials and the left hand side becomes e^x and the digamma function for integers arguments can be expressed through harmonic numbers and the euler mascheroni constant:

$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1) =\ln(x)e^x\\\\

\ln(x) = e^{-x}(1 - \gamma + \sum_{n=1}^{\infty} \frac{x^n}{n!} (\sum_{k=1}^{n} \frac{1}{k} - \gamma))$ where $\displaystyle \gamma$ is the euler mascheroni constant.

I decided to multiply the infinite series and I got:

$\displaystyle ln(x) = \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\sum_{c=1}^{n-k}\frac{1}{c} - \gamma}{k!(n-k)!})$

I can't write code for this in pari however, it's too complex. So Instead I just used the first series and multiplied it by $\displaystyle e^{-x}$.

I've been unable to properly do the ratio test but using Pari gp it seems to converge for values x >e, but failed at 1000, worked for 900.

Any questions comments would be appreciated, thanks.

Specifically helpwise, I was wondering if anyone knows how I can prove the convergence? And if this is maybe old news and I'm proving something that already exists? And I'd like to know if this is actually novel and something worth proving, thanks.