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

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.

• What do you mean by "proof is available for $x\in\{0,1\}$"? – Nicco Oct 28 '15 at 12:11
• What I mean is that Mathematica knows it is true in those cases. – Nazgand Oct 28 '15 at 12:43
• This would be a very nice result to establish. Could you tell us more concerning the way you have found out this possible equality ? – Tom-Tom Oct 28 '15 at 12:49
• @Tom-Tom Sure. 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 $\frac{11-4 e}{2 e-5}$. – Nazgand Oct 28 '15 at 13:16
• The closed form matched all digits. I used Wolfram Alpha again to find out N[ContinuedFractionK[n, n + 3, {n, 1, 10000}], 190] suggested $(49-18 E)/(2 (3 E-8))$ 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 $-x$. – Nazgand Oct 28 '15 at 13:30

O. Perron, Die Lehre von den Kettenbrüchen, Chapter XI, Section 81.

Here is formula (8), p. 477 (in the 2nd edition, 1922) $$\gamma-x + \underset{n=1}{\overset{\infty }{\mathbf K}}\; \frac{(\beta+n)x}{\gamma-x+n} = \frac{{}_1F_1(\beta,\gamma,x)\;\gamma}{{}_1F_1(\beta+1,\gamma+1,x)} \tag{8}$$ $(x \ne 0, \beta\ne -1, -2, -3, \dots)$. The reference is .

Making the appropriate substitutions, we get yours as $$x+\underset{n=1}{\overset{\infty }{\mathbf K}}\; \frac{n}{x+n} = \frac{{}_1F_1(0,x+1,1)\;(x+1)}{{}_1F_1(1,x+2,1)} =\frac{(x+1)}{{}_1F_1(1,x+2,1)} .$$ Evaluate the series ${}_1F_1(1,x+2,1)$ to get your answer.

 O. Perron, "Über eine spezielle Klasse von Kettenbrüchen". Rend. Pal. 29 (1910)

• FullSimplify[ 1/E/(Gamma[x + 1, 0, 1]) == (x + 1)/Hypergeometric1F1[1, x + 2, 1]] results as true in Mathematica, so I will mark your answer as the correct one when I understand the proof for the (8) formula you posted. I could not find a complete copy of "Über eine spezielle Klasse von Kettenbrüchen", and thus still do not have access to the proof I desire. – Nazgand Oct 28 '15 at 16:08
• @Nicco The book "Die Lehre von den Kettenbrüchen" is here: books.google.com/… – Nazgand Oct 28 '15 at 17:37
• @Nazgand:congratulations for rediscovering something that already existed.I encourage you to keep on looking for such interesting patterns. – Nicco Oct 28 '15 at 18:01
• @Nicco is right. Keep exploring. – GEdgar Oct 28 '15 at 18:08
• I found a copy of "Über eine spezielle Klasse von Kettenbrüchen". It is here: dropbox.com/s/7qtzsb8vgfxdgje/art3A10.10072FBF03014063.pdf?dl=0 I shall attempt to translate it and use it to prove my result. By the way, do you think arxiv would accept me publishing my translation? – Nazgand Oct 28 '15 at 18:23