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While working on an earlier question involving $\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}}$ I rewrote the sum as a contour integral, using generating functions: $$ \sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} = \sum_{j=0}^n \left( \frac{1}{4^n} \binom{n+j-1}{j} \right) 2^{n-j} = [t^n] \left( \sum_{j=0}^\infty \frac{t^j}{4^n}\binom{n+j-1}{j} \cdot \sum_{j=0}^\infty t^j 2^j \right) $$ Now, using well known generating functions, for $|t|<1/2$: $$ \sum_{j=0}^\infty t^j \binom{n+j-1}{j} = \frac{1}{(1-t)^n} \quad \sum_{j=0}^\infty (2t)^j = \frac{1}{1-2t} $$ We get $$ \sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} = [t^n] \left( \frac{1}{1-2t} \frac{1}{\left(4(1-t)\right)^n} \right) = \frac{1}{2 \pi i} \oint \frac{1}{1-2t} \left(\frac{1}{4 t(1-t)} \right)^n \frac{\mathrm{d} t}{t} $$ where the Cauchy integral formula was used along with $n! [t^n] g(t) = g^{(n)}(0)$.

Now, Byron Schmuland showed that the large $n$ limit of the left-hand-side equals $\frac{1}{2}$.

Question: Can one demonstrate $$ \lim_{n \to \infty} \frac{1}{2 \pi i} \oint_{ |t| = \rho} \frac{1}{1-2t} \left(\frac{1}{4 t(1-t)} \right)^n \frac{\mathrm{d} t}{t} = \frac{1}{2}$$ using asymptotic methods? Here $0<\rho<\frac{1}{2}$.

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Sasha
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1 Answers1

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As $t$ traces around the origin, so does $u=4 t (1-t)$. Solving for $t$ we get two solutions, one, $2t=1-\sqrt{1-u}$ maps a $u$-path around the origin into the $t$-path around the origin, and another, $2t=1+\sqrt{1-u}$, maps a $u$-path around the origin into the $t$-path around $t=1$. This suggests a change of variables, $t = \frac{1}{2} \left(1-\sqrt{1-u}\right)$ $$ \frac{1}{2 \pi i} \oint \frac{1}{\left(4 t(1-t) \right)^n} \frac{\mathrm{d}t}{t(1-2t)} = \frac{1}{2 \pi i} \oint \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u} \frac{\mathrm{d}u}{u^{n+1}} = [u^n] \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u} $$ where the Cauchy formula was used in the last equality. This gives $$\begin{eqnarray} \sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} &=& [u^n] \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u} = [u^n] \frac{1}{2} \left( \frac{1}{1-u} + \frac{1}{\sqrt{1-u}} \right) \\ &=& \frac{1}{2} + \frac{1}{2} \binom{-1/2}{n} = \frac{1}{2} + \frac{1}{2^{2n+1}} \binom{2n}{n} \end{eqnarray} $$ Although this is not quite what you are asking for, this reproduces Byron's closed form result leading to the evaluation of the limit.

Alternatively, you could infer the large $n$ asymptotioc from the near-singularity-behavior of the generating function $g(u) = \frac{1}{2} \frac{1+ \sqrt{1-u}}{1-u}$: $$ g(u) = \frac{1}{2} \frac{1}{1-u} + \mathcal{o}\left(\frac{1}{1-u}\right) $$ implying the limit equals $[u^n] g(u) \sim \frac{1}{2}$.

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