Continued fractions II

Continued fractions for confluent hypergeometric functions

Today we show the proof of this nice continued fraction proposed by @nntaleb \[ \frac{x}{e^x -1}= 1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3x}{6 + \cdots}}}}} \] The proof will relly on the continued fractions theorem for confluent hypergeometric functions

Proof

Note that \[ e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{(1)_{n}}{(1)_{n}}\frac{x^n}{n!} = {}_{1}F_{1} (1;1;x) \] where \[(x)_{n} = x(x+1)(x+2)\cdots(x+n-1)\] is the Pochhhammer polynomial or rising factorial of degree $n$ and ${}_{1}F_{1}(a;b;z)$ is the confluent hypergeometric function, also denoted by $M(a;b;z)$. There is a continued fractions theorem for confluent hypergeometric functions:

Thorem Consider the function:

\[ {}_{1}F_{1}(b;c;z)= \sum_{n=0}^{\infty} \frac{(b)_{n}}{(c)_{n}} \frac{z^n}{n!}\] Let $\left\{a_{n}\right\}$ be a sequence of complex numbers defined by \[ a_{2n} = \frac{b+n}{(c+2n-1)(c+2n)} \quad n=1,2,3,...\] \[ a_{2n+1} = -\frac{c-b+n}{(c+2n)(c+2n+1)} \quad n=0,1,2,3,...\] Then \[ 1+K_{n=1}^{\infty}\left(\frac{a_{n}z}{1}\right) = \frac{{}_{1}F_{1} (b;c;z)}{{}_{1}F_{1} (b+1;c+1;z)} \quad \forall z\in \mathbb{C} \]

Corollary If we let $b\to 0$ and making the transformation $c+1 \mapsto c$

\[ 1-\cfrac{\frac{1}{c}z}{1+\cfrac{\frac{1}{c(c+1)}z}{1-\cfrac{\frac{c}{(c+1)(c+2)}z}{1+\cfrac{\frac{2}{(c+2)(c+3)}z}{1-\cfrac{\frac{c+1}{(c+3)(c+4)}z}{1+ \cfrac{\frac{3}{(c+4)(c+5)}z}{1-\cdots}}}}}}= \cfrac{1}{{}_{1}F_{1} (1;c;z) } \] \begin{align*} \Longrightarrow {}_{1}F_{1} (1;c;z) =&\cfrac{1}{1-\cfrac{\frac{1}{c}z}{1+\cfrac{\frac{1}{c(c+1)}z}{1-\cfrac{\frac{c}{(c+1)(c+2)}z}{1+\cfrac{\frac{2}{(c+2)(c+3)}z}{1-\cfrac{\frac{c+1}{(c+3)(c+4)}z}{1+ \cfrac{\frac{3}{(c+4)(c+5)}z}{1-\cdots}}}}}}}\\ =& \cfrac{1}{1-\cfrac{z}{c+\cfrac{z}{c+1 -\cfrac{cz}{c+1+\cfrac{2z}{c+3-\cfrac{(c+1)z}{c+4+\cfrac{3z}{c+5 - \cdots}}}}}}} \tag{*} \end{align*} Applying $(*)$ to $ e^{x}$: \[ e^x = {}_{1}F_{1}(1;1;x) = \cfrac{1}{1-\cfrac{x}{1+\cfrac{x}{2 -\cfrac{x}{3+\cfrac{2x}{4-\cfrac{2x}{5+\cfrac{3z}{6 - \cdots}}}}}}} \] Making the transformation $x\mapsto -x$: \[ \Longrightarrow e^{-x} = \frac{1}{e^x} = \cfrac{1}{1+\cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}}} \] \[ \Longrightarrow e^x = 1+\cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}} \] \[ \Longrightarrow e^x -1 = \cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}} \] Finally, taking the reciprocal: \[\boxed{ \Longrightarrow \frac{x}{e^x -1}= 1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}} } \]