How can we write the expansion of $\frac{e^t(e^t-1)^{n-1}}{(e^t+1)^{n+1}}.$

Question

How can we write the expansion of $$\frac{e^t(e^t-1)^{n-1}}{(e^t+1)^{n+1}}.$$

I know that $$\frac{1}{(e^t+1)^{n+1}}=\sum_{k=0}^\infty (-1)^k \binom{n+k}{k} e^{kt}.$$

Could you please give me an idea?

Answer 1

This is what Wolfram alpha produces (https://www.wolframalpha.com/input/?i=series+e%5Et(e%5Et-1)%5E%7Bn-1%7D%7D%2F%7B(e%5Et%2B1)%5E%7Bn%2B1%7D):

Answer 2

If you wish to perform a series expansion in terms of $e^t$, then define $$f(z) = \frac{z(z-1)^{n-1}}{(z+1)^{n+1}}$$ and your desired expansion corresponds to the expansion of $f(z)$ with respect to $z = 0$, followed by the substitution $z = e^t$.

Next, observe $$\sum_{k=n}^\infty (-1)^k \frac{k!}{(k-n)!} z^{k-n} = \frac{d^n}{dz^n}\left[\frac{1}{z+1}\right] = \frac{(-1)^n n!}{(z+1)^{n+1}},$$ so that $$\frac{1}{(z+1)^{n+1}} = \sum_{k=n}^\infty \binom{k}{n} (-z)^{k-n} = \sum_{k=0}^\infty \binom{k+n}{n} (-z)^k.$$ This can of course be found from the binomial series. Consequently, with the convention that $\binom{a}{b} = 0$ if $a < b$, and the transformation of indices $s = k+m$, $j = m$, $$\begin{align*} f(z) &= z \sum_{m=0}^{n-1} \binom{n-1}{m} z^m (-1)^{n-1-m} \sum_{k=0}^\infty \binom{k+n}{n} (-z)^k \\ &= z \sum_{k=0}^\infty \sum_{m=0}^{n-1} \binom{n-1}{m} \binom{k+n}{n} (-1)^{n-1-m+k} z^{k+m} \\ &= \sum_{s=0}^\infty \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{s-j+n}{n} (-1)^{n+s-2j-1} z^{s+1}. \\ \end{align*}$$ The coefficient for $z^{s+1}$, which we define as $$c(s) = \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{s-j+n}{n} (-1)^{n+s-2j-1},$$ is expressible as a hypergeometric function but doesn't appear to have a simple closed form for general $n$.

Answer 3

I think that writing the general term could be difficult.

For a truncated series, what I would do considering $$y=\frac{e^t(e^t-1)^{n-1}}{(e^t+1)^{n+1}}$$ is to take logarithms of both sides and expand as Taylor series at $t=0$. This should give $$\log(y)=(n-1) \log (t)+\log \left(2^{-n-1}\right)-\frac{(n+2)}{12} t^2+\frac{(7 n+8) }{1440}t^4-\frac{(31 n+32) }{90720}t^6+\frac{(127 n+128) }{4838400}t^8-\frac{(511 n+512) }{239500800}t^{10}+O\left(t^{12}\right)$$ and continue with Taylor series using $y=e^{\log(y)}$ and here start the complexity since we should arrive to something looking like $$\frac{y}{t^{n-1}}=2^{-(n+1)}+\sum_{k=1}^p (-1)^k \alpha_k\,t^{2k} $$ with $$\alpha_1=\frac{2^{-(n+3)}}{3} (n+2)$$ $$\alpha_2=\frac{2^{-(n+6)}}{45} (n+4) (5 n+7)$$ $$\alpha_3=\frac{2^{-(n+8)}}{2835} (n+6)(35 n^2+147 n+124)$$ $$\alpha_4=\frac{2^{-(n+12)}}{42525} (n+8)(175 n^3+1470 n^2+3509 n+2286)$$ $$\alpha_5=\frac{2^{-(n+14)}}{1403325} (n+10)(385 n^4+5390 n^3+24959 n^2+44242 n+24528)$$ Another possible approach would be to write

$$y=\frac{1}{4} \text{sech}^2\left(\frac{t}{2}\right)\tanh ^n\left(\frac{t}{2}\right)$$ and use $$\tanh \left(\frac{t}{2}\right)=2\sum_{k=1}^\infty \frac{ \left(4^k-1\right) B_{2 k} }{(2 k)!}t^{2 k-1}$$ $$\text{sech}\left(\frac{t}{2}\right)=\sum_{k=0}^\infty \frac{ E_{2 k} }{4 ^k\,(2 k)!}t^{2 k}$$ which give, for $y$, an expression that .... you just need to expand !