Solve for $(2t-1)^{-1}$ in binomial and hypergeometric form

Question

Solve for $(2t-1)^{-1}$ in binomial and hypergeometric form. We have from the binomial theorem $$(a+x)^k=\sum_{n=0}^{\infty}\binom{k}{n}a^{k-n}x^n$$ For this instance, $a=2t, x=-1, k=-1$ $$(2t-1)^{-1}=\lim_{m \to \infty}\sum_{n=0}^{m}\frac{(-1)!}{n!}(2t)^{-1-n}(-1)^n$$ I have that $(-1)!=(-1)^n(1)_n$, $$(2t-1)^{-1}=\lim_{m \to \infty}\frac{1}{2t}\sum_{n=0}^m\frac{(1)_n}{n!}\left(\frac{1}{2t}\right)^n=\frac{\left(\sqrt{e}\right)^{t^{-1}}}{2t}$$

Which produces the following hypergeometric function $_2F_1(1;-;-;\frac{1}{2t})$

The equality above does not work for $t=1$ and so I question whether this approach is correct?

Answer 1

We expand the rational function $\frac{1}{2t-1}$ at $t=0$.

The binomial expansion at $t=0$ gives \begin{align*} \color{blue}{\frac{1}{2t-1}}&=-(1-2t)^{-1} =-\sum_{n=0}^{\infty}\binom{-1}{n}(-2t)^n\tag{1}\\ &\,\,\color{blue}{=-\sum_{n=0}^{\infty}(2t)^n}\tag{2} \end{align*} which corresponds with the expansion as geometric series.

In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ so that $ \binom{-1}{n}=\binom{n}{n}(-1)^n=(-1)^n $.

From (2) we derive a representation as hypergeometric series: \begin{align*} \color{blue}{\frac{1}{2t-1}}& =-\sum_{n=0}^{\infty}(2t)^n =-\sum_{n=0}^{\infty}n!\frac{(2t)^n}{n!}\\ &=-\sum_{n=0}^{\infty}(1)_n\frac{(2t)^n}{n!}\\ &\,\,\color{blue}{=-{}_1F_0(1;0;2t)} \end{align*}