# Continued Fraction: Please prove $\frac{1}{e \gamma (x+1,1)}=x+\frac{1}{x+1+\frac{2}{x+2+\frac{3}{x+3+\frac{4}{\dots}}}}$ [closed]

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 HumphriesOct 28 '15 at 18:17

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