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

Solution 1:

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)