# Asymptotics of an almost binomial sum

Take some $$d,C\in\mathbb{C}$$ and let

$$A_{n}=\sum_{m=0}^{n}\binom{n}{m}\left(1+\frac{m}{n}\right)^{d}C^{m}.$$

I want to calculate the asymptotics as $$n$$ tends to $$\infty$$. Numerically in a few examples and proven for $$d\in\mathbb{Z}_{\geq0}$$ it has leading order given by

$$A_{n}\sim (C+1)^{n}\left(1+\frac{C}{C+1}\right)^{d}.$$

My proof for $$d\in\mathbb{Z}_{\geq0}$$ used

$$A_{n}=(C+1)^{n}\sum_{k=0}^{n}\binom{d}{k}\left(\frac{C}{C+1}\right)^{k}\sum_{r=0}^{\infty}\frac{\binom{d-k}{r}(n)_{k}S(r+k,k)}{\binom{r+k}{r}n^{r+k}}$$

where $$S(r+k,k)$$ are the Stirling numbers of the second kind and as $$d\in\mathbb{Z}_{\geq0}$$ the sums are just over $$k=0,...,d$$ and $$r=0,...,d-k$$.

Any ideas would be great! Thanks in advance!

If you take out a factor $$(C+1)^n$$, this is

$$(C+1)^n\sum_{m=0}^n\binom nm\left(1+\frac mn\right)^d\left(\frac C{C+1}\right)^m\left(\frac1{C+1}\right)^{n-m}\;.$$

Setting $$p=\frac C{C+1}$$ and $$1-p=\frac1{C+1}$$ shows that this is an expectation with a “complex-valued binomial distribution”:

$$(C+1)^n\sum_{m=0}^n\left(1+\frac mn\right)^d\binom nmp^m(1-p)^{n-m}\;.$$

With increasing $$n$$, the distribution gets concentrated around the “mean” $$np$$, so we get a first approximation by substituting $$p=\frac C{C+1}$$ for $$\frac mn$$:

$$(C+1)^n\left(1+\frac C{C+1}\right)^d\;.$$

This can be used to systematically derive the asymptotic expansion in inverse powers of $$n$$ by expanding the “random variable” in powers of the deviation from the mean:

$$\begin{eqnarray} \left(1+\frac mn\right)^d &=&\left(1+p+\frac mn-p\right)^d \\ &=& (1+p)^d\sum_{k=0}^\infty\binom dk\left(\frac{m-np}{n(1+p)}\right)^k\;, \end{eqnarray}$$

which yields the sum in terms of the central moments $$\mu_k$$ of the binomial distribution:

$$(C+1)^n(1+p)^d\sum_{k=0}^d\binom dk(n(1+p))^{-k}\mu_k(n,p) \\ = (C+1)^n\left(1+\frac C{C+1}\right)^d\sum_{k=0}^d\binom dk\left(n\left(1+\frac C{C+1}\right)\right)^{-k}\mu_k\left(n,\frac C{C+1}\right)\;.$$

The first six central moments of the binomial distribution are given at Wikipedia, and a recurrence relation for them is derived in Closed-Form Expressions for the Moments of the Binomial Probability Distribution by Andreas Knoblauch. As might be expected from the linear increase of the variance with $$n$$, the $$k$$-th central moment grows with the $$\left\lfloor\frac n2\right\rfloor$$-th power of $$n$$, so that we need two central moments to get one inverse power of $$n$$ in the expansion. Up to $$O \left(n^{-2}\right)$$, with the well-known central moments $$\mu_0=1$$, $$\mu_1=0$$ and $$\mu_2=np(1-p)$$, the result is

$$(C+1)^n\left(1+\frac C{C+1}\right)^d\left(1+\frac1n\binom d2\frac C{(2C+1)^2}+O\left(\frac1{n^2}\right)\right)\;.$$

Note that while the binomial distribution is usually only used for $$p\in[0,1]$$, the central moments are purely algebraic quantities that have the same algebraic form for any $$p\in\mathbb C$$.

• Hey thanks! Does the argument still work when C/(C+1) is not in the interval [0,1]? As I guess you approximate the delta(x-C/(C+1)) and then only integrate over [0,1] right? May 7, 2020 at 8:57
• @Campbell: Sorry, I'd forgotten that you explicitly wrote that $C\in\mathbb C$. The argument didn't really work for complex values the way I'd formulated it – one would have had to take into account a complex phase in the Gaussian – I completely rewrote the answer based on the central binomial distribution, which not only makes for a cleaner argument but also yields the entire asymptotic expansion in inverse powers of $n$. May 7, 2020 at 18:15

Just another idea: $$\left(1+\frac{m}{n}\right)^d=\frac{\partial^d}{\partial x^d}\Big[e^{(1+m/n)x}\Big]_{x=0}\implies(1+C)^{-n}A_n=\frac{\partial^d}{\partial x^d}\left[e^x\left(\frac{1+Ce^{x/n}}{1+C}\right)^n\right]_{x=0}.$$