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

A: 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 !
