# Power series identity [duplicate]

Through a numerical computation, I stumbled across the following identity. It takes place in the ring $(\mathbb{Z}[x])[[t]]$, which is complete with respect to the $t$-adic valuation. The apparent identity is $$\sum_{n=0}^\infty\frac{x(x+1)(x+2)\cdots(x+n-1)}{(1+t)(1+2t)\cdots(1+nt)}t^n=\frac1{1-xt}$$

I have numerically verified that this holds $\bmod t^{50}$. Does anyone have any ideas about how to prove this identity?

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## marked as duplicate by J. M., William, rschwieb, Emily, BenjaLim Sep 23 '12 at 23:37

Your identity is a special case of Gauss's hypergeometric identity (see here for a proof),

\begin{align*} \sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} (1+(j+1)t)}t^n&=\sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} \left(1+\frac1{t}+j\right)}\\ &=\sum_{n=0}^\infty \frac{(x)_n}{\left(1+\frac1{t}\right)_n}=\sum_{n=0}^\infty \frac{(x)_n (1)_n}{\left(1+\frac1{t}\right)_n}\frac1{n!}\\ &={}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\Gamma\left(1+\frac1{t}\right)\Gamma\left(\frac1{t}-x\right)}{\Gamma\left(\frac1{t}\right)\Gamma \left(1+\frac1{t}-x\right)} \end{align*}

where ${}_2 F_1\left({{a,b}\atop{c}}\mid z\right)$ is the Gaussian hypergeometric function, and since $\Gamma(1+z)=z\Gamma(z)$,

$$\require{cancel} {}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma\left(\frac1{t}-x\right)}}{t\left(\frac1{t}-x\right)\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma \left(\frac1{t}-x\right)}}=\frac1{t\left(\frac1{t}-x\right)}=\frac1{1-xt}$$

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