I have been playing around with Mathematica and continued fractions and I noticed something.

ContinuedFractionK[n, n + x, {n, 1, Infinity}] ==-x + 1/(E Gamma[1 + x] - E Gamma[1 + x, 1])==-x + 1/(E Gamma[1 + x, 0, 1])

$$\underset{n=1}{\overset{\infty }{K}}\frac{n}{n+x}=\frac{1}{e \gamma (x+1,1)}-x$$

In more traditional notation, this means

$$\frac{1}{e \gamma (x+1,1)}=x+\frac{1}{x+1+\frac{2}{x+2+\frac{3}{x+3+\frac{4}{\dots}}}}$$

I have verified this to be true for all x I have tested including complex numbers for hundreds of digits. I do not know how to prove my result and would like proof. Mathematica can verify $x\in\{0,1\}$. Failing proof of all x, proof with x being another specific number e.g. 2 gets partial credit.

PS. $\gamma (a,b)$ is the lower incomplete gamma function.

Note: For whole numbered x, we have $$\frac{1}{e*x!-A000522(x)}=x+\frac{1}{x+1+\frac{2}{x+2+\frac{3}{x+3+\frac{4}{\dots}}}}$$ https://oeis.org/A000522

This is the story of my discovery: I noticed that ContinuedFractionK[n, n, {n, 1, Infinity}] was $\frac{1}{e-1}$ and ContinuedFractionK[n, n + 1, {n, 1, Infinity}] was $\frac{1}{e-2}-1$. Because the answers were both fractions of e, I thought I could find a pattern. ContinuedFractionK[n, n + 2, {n, 1, Infinity}] and onward gave no results, but I didn't let that stop me. I calculated N[ContinuedFractionK[n, n + 2, {n, 1, 10000}], 190] for an approximation and pasted the first 190 digits into Wolfram Alpha where it suggested the possible closed form of $\frac{11-4 e}{2 e-5}$. The closed form matched all digits. I used Wolfram Alpha again to find out N[ContinuedFractionK[n, n + 3, {n, 1, 10000}], 190] suggested $\frac{49-18 e}{2 (3 e-8)}$ and that N[ContinuedFractionK[n, n + 4, {n, 1, 10000}], 190] suggested $-\frac{3 (32 e-87)}{24 e-65}$. Wolfram Alpha failed to be of further help with closed forms, so I made a list of what I knew, used the Expand[] and FullSimplify[] functions on it to get: $\left\{\frac{1}{e-1},\frac{1}{e-2}-1,\frac{1}{2 e-5}-2,\frac{1}{6 e-16}-3,\frac{1}{24 e-65}-4\right\}$. The numbers by the e's were obviously $x!$, outside the fraction was $−x$. This left the numbers {1,2,5,16,65}. A lookup found this: https://oeis.org/A000522 with the formula a(n) = e*Gamma(n+1,1) I thus conjectured that ContinuedFractionK[n, n + x, {n, 1, Infinity}]$=\frac{1}{e \Gamma (x+1)-e \Gamma (x+1,1)}-x$. Every x I have tested on the complex plane has shown this to be correct.


closed as unclear what you're asking by Alexey Ustinov, Myshkin, Boris Bukh, Will Jagy, Peter Humphries Oct 28 '15 at 18:17

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    $\begingroup$ Continued fraction representations for incomplete gamma function functions.wolfram.com/GammaBetaErf/Gamma2/10 $\endgroup$ – Alexey Ustinov Oct 28 '15 at 14:17
  • $\begingroup$ @AlexeyUstinov That is good. Thanks. Perhaps someone will see how to manipulate one of those to match my continued fraction. $\endgroup$ – Nazgand Oct 28 '15 at 14:25