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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.

Nazgand
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  • 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
  • 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. – Nazgand Oct 28 '15 at 13:33
  • Thanks a lot for the story. This kind of information is quite useful. In my opinion, it is worth being included in the question. – Tom-Tom Oct 28 '15 at 13:52
  • Continued fraction representations for incomplete gamma function http://functions.wolfram.com/GammaBetaErf/Gamma2/10/ Maybe someone could manipulate one of them into my continued fraction? – Nazgand Oct 28 '15 at 14:22
  • See Euler's continued fraction formula and Gauss's continued fraction. – Lucian Oct 28 '15 at 19:57

1 Answers1

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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 [1].

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.

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

GEdgar
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  • 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: https://books.google.com/books?id=Mb8SAwAAQBAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false – Nazgand Oct 28 '15 at 17:37
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    @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
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    @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: https://www.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
  • @Nazgand:It's best to give it a try,you never know. – Nicco Oct 31 '15 at 08:01
  • I still haven't translated enough to understand the proof, but I am pretty sure I have the right info, so I will just mark this as the answer. – Nazgand Nov 23 '15 at 02:49