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I am trying to answer this question. Per suggestion in one the comments for that question, one might be able use the power series of the terms to arrive at the answer. However, one of the terms is:

$$\left(\frac{1+x}{1-x}\right)^n$$

I am wondering if there is an explicit power series formula for this expression when $x$ is small.

I cannot see a pattern in the expression for the Taylor series output by Mathematica:

$\left(\frac{1+x}{1-x}\right)^n=1+2 n x+2 n^2 x^2+\frac{2}{3} \left(2 n^3+n\right) x^3+\frac{2}{3} \left(n^4+2 n^2\right) x^4+\frac{2}{15} \left(2 n^5+10 n^3+3 n\right) x^5+\frac{2}{45} \left(2 n^6+20 n^4+23 n^2\right) x^6+\frac{2}{315} \left(4 n^7+70 n^5+196 n^3+45 n\right) x^7+\ldots$

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M.B.M.
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2 Answers2

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Let us attack the top and the bottom separately and then combine them. By the binomial theorem, $(1+x)^n=\sum \binom{n}{k} x^k$. Note, I am taking binomial coefficients to be zero unless both the top and bottom terms are non-negative integers.

The bottom is $\frac{1}{(1-x)^n}=(1+x+x^2+\ldots)^n$. By using the same reasoning as in the combination interpretation of the [multinomial theorem], we have that the coefficient of $x^k$ will be the number of ways of writing $x^k$ as a product of $x^{a_i}$ with $1\leq i\leq n$ and $0\leq a_i$. This, in turn, is the number of ways of writing $k=\sum_{i=1}^n a_i$, which is the number of partitions of $k$ into $n$ (possibly of size zero nonzero) pieces. By stars and bars, this will be $\binom{k+(n-1)}{k}$.

Power series multiply by convaluting the coefficients. Therefore, $$\left(\frac{1+x}{1-x} \right)^n=\left(\sum_k \binom{n}{k}x^k \right)\left( \sum_k\binom{(n-1)+k}{k}x^k \right)=\sum_k \left(\sum_{i+j=k,0\leq i,j} \binom{n}{i} \binom{(n-1)+j}{j}\right)x^k .$$

There may be a way to simplify the convolution further, but I do not see it off the top of my head.

Aaron
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I have likely made a small mistake in this gross derivation, but this is an idea:

$\displaystyle \Big(\frac{1+x}{1-x}\Big)^n=\frac{1}{(1-x)^n}(1+x)^n=(1+x)^n\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\frac{1}{1-x}=(x+1)^n\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\sum_{i=0}^\infty x^i=\frac{1}{(n-1)!}\sum_{k=0}^n {n\choose k} x^{n-k}\sum_{i=0}^\infty \frac{(i+(n-1)!}{(i)!}x^{i}=\frac{1}{(n-1)!}\sum_{k=0}^\infty {n\choose k}\chi_{k\le n} x^{n-k}\sum_{i=0}^\infty \frac{(i+(n-1)!}{(i)!}x^{i}=\sum_{l=0}^\infty\Bigg(\frac{1}{(n-1)!}\sum_{m=0}^l{n\choose m}\chi_{m\le n}\frac{(l-m-(n-1)!}{(l-m)!}\Bigg)x^l$

where the last part uses the Cauchy product. Note that $\chi_{m\le n}=1$ if $m\le n$, $0$ otherwise. You get negative factorials, but that's why I used the indicator function to make clear that those terms are $0$.

JLA
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  • Thanks, but I think you flipped plus to minus in the numerator after the first equality (i.e. it should be $\frac{1}{(1-x)^n}(1+x)^n$...) Still checking whether this impacts the rest of derivation... – M.B.M. Aug 11 '13 at 20:55
  • Oh you are right, I will fix that. Shouldn't change much. – JLA Aug 11 '13 at 20:58
  • You don't have the $(-1)^{n-1}$. Differentiating $(1-x)^{-k}$ produces two minus signs, one from the exponent, and one from the $1-x$, they cancel each other. – Daniel Fischer Aug 11 '13 at 22:22
  • Woops. Corrected. – JLA Aug 11 '13 at 22:41