Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0<\rho<\frac{1}{2}$.

share|cite|improve this question

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 $[u^n] g(u) \sim \frac{1}{2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.