# Does a series of binomial coefficients converge?

Define $$\sigma_m = \sum_{j=0}^{m-1} \left| \binom{1/2}{j} \right|$$ Does $\lim_{m \to \infty} \sigma_m$ converge? Here is a plot of the first $64$ sums:

A plot of the first 1024 elements seem to suggest that $\sigma_m \to 2$. (Can be seen by invoking Abel's Theorem on $(1+x)^{1/2}$) Observe that $$\left| \frac{\binom{1/2}{j+1}}{\binom{1/2}{j}} \right| = \left|\frac{1/2 - j}{j+1}\right| \to 1$$ so the ratio test is inconclusive.

I also tried expanding out the terms: $$\left| \binom{1/2}{j} \right| = \binom{2j}{j} \frac{1}{2^{2j} (2j-1)} = \frac{(2j)!}{(j!)^2 2^{2j} (2j-1)}$$

This problem arises when showing the analyticity of $(1-x)^{1/2}$.

Let $(\alpha)_n = \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)} = \alpha (\alpha+1)\cdots(\alpha+n-1)$ be the rising factorials. Notice $$\frac{(\alpha)_n}{n!} - \frac{(\alpha)_{n-1}}{(n-1)!} = (( \alpha + n - 1 ) - n )\frac{(\alpha)_{n-1}}{n!} = \frac{(\alpha-1)_n}{n!}$$ Substitute $\alpha = \frac12$ and compare the expansion of $\displaystyle\;\frac{(-\frac12)_n}{n!}$ against that of $\displaystyle\;\binom{\frac12}{n}$. We find for $n > 0$.

$$\frac{(\frac12)_n}{n!} - \frac{(\frac12)_{n-1}}{(n-1)!} = \frac{(-\frac12)_n}{n!} = - \left|\binom{\frac12}{n}\right|$$

From this, we find for $m > 1$,

$$\sigma_m = \sum_{j=0}^{m-1}\left|\binom{\frac12}{j}\right| = 1 + \sum_{j=1}^{m-1}\left[\frac{(\frac12)_{j-1}}{(j-1)!} - \frac{(\frac12)_j}{j!}\right] = 2 - \frac{(\frac12)_{m-1}}{(m-1)!}$$ Notice the term $\displaystyle\;\frac{(\frac12)_{m-1}}{(m-1)!} = \prod_{k=1}^{m-1}\frac{k-\frac12}{k} = \prod_{k=1}^{m-1}\left(1-\frac{1}{2k}\right) \;$ is clearly positive and we can bound it from above by $$\prod_{k=1}^{m-1}\exp\left(-\frac{1}{2k}\right) = \exp\left(-\frac12\sum_{k=1}^{m-1}\frac1k\right) \le \exp\left(-\frac12\sum_{k=1}^{m-1}\int_{k}^{k+1} \frac{dx}{x}\right) = \frac{1}{\sqrt{m}}$$ This means $2 - \sigma_m = O\left( \frac{1}{\sqrt{m}} \right)$ and hence $\lim\limits_{m\to\infty} \sigma_m = 2$.

For a better estimate of $2 - \sigma_m$, we can use the fact $\displaystyle\;\lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)n^\alpha} = 1\;$ for all $\alpha \in \mathbb{R}$.
This allow us to fix the coefficient in front of $\frac{1}{\sqrt{m}}$ behavior: $$2 - \sigma_m = \frac{(\frac12)_{m-1}}{(m-1)!} = \frac{\Gamma(m-\frac12)}{\Gamma(\frac12)\Gamma(m)}\approx \frac{1}{\sqrt{\pi m}}$$

The following test is valid: If for all $n>n_0$ and some $\alpha>1$ $$|a_{n+1}|\leq\left( 1-\frac\alpha{m}\right) |a_n|$$ then $\sum |a_n|$ converges.

This is true because $$\sum_{k=a+1}^\infty \prod_{m=a+1}^k \left( 1-\frac\alpha{m}\right) = \frac{\Gamma(a-1)\Gamma(a+2)}{(a+1)\Gamma(a+1)\Gamma(a)}$$ and the few remaining terms for $k<a+1$ at any rate have a finite sum.

This test works on your problem. BTW, you are right, the sum converges to $2$.

• Should the first $m$ be $n$? – Henricus V. Nov 10 '16 at 1:49
• How can you prove the equality without using $\Gamma$ function identities? – Henricus V. Nov 10 '16 at 5:20

Hint: Stirling shows that $(2j)!/(j!)^2$ is asymptotic to a constant times $2^{2j}j^{-1/2}.$


By using the Euler Reflection Formula we can show that $${1/2 \choose j} = {\color{#f00}{\pars{-1}^{j + 1}} \over 2\root{\pi}} \,{\Gamma\pars{j - 1/2} \over j!}\qquad\mbox{and}\qquad {1/2 \choose 0} = 1.$$ Note that for $\ds{\color{#f00}{j \geq 1}}$, $\ds{\quad{1/2 \choose j}_{\ j\ \mrm{even}} < 0}$ and $\ds{\quad{1/2 \choose j}_{\ j\ \mrm{odd}} > 0}$

Then, \begin{align} \sum_{j = 0}^{\infty}\verts{1/2 \choose j} & = 1 + \sum_{j = 0}^{\infty}\bracks{{1/2 \choose 2j + 1} - {1/2 \choose 2j + 2}} \\[5mm] & = 1 + \sum_{j = 0}^{\infty}\bracks{{1/2 \choose j + 1} - {1/2 \choose j + 2}} {1 + \pars{-1}^{\, j} \over 2} \\[5mm] & = 1 + {1 \over 2}\sum_{j = 0}^{\infty}{1/2 \choose j + 1}\bracks{1 + \pars{-1}^{\, j}} - {1 \over 2}\sum_{j = 0}^{\infty}{1/2 \choose j + 1}\bracks{1 - \pars{-1}^{\, j}} \\[5mm] & = 1 + \sum_{j = 0}^{\infty}{1/2 \choose j + 1}\pars{-1}^{\, j} = 1 - \sum_{j = 1}^{\infty}{1/2 \choose j}\pars{-1}^{\, j} \\[5mm] & = 1 - \braces{\bracks{1 + \pars{-1}}^{1/2} - {1/2 \choose 0}} = \bbox[#ffe,15px,border:2px dotted navy]{\ds{2}} \end{align}