Xrfgtjtk

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?

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 $n$.

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):

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
$$log(y)=(n-1) log (t)+log left(2^{-n-1}right)-frac{(n+2)}{12}
t^2+frac{(7 n+8) }{1440}t^4-frac{(31 n+32) }{90720}t^6+frac{(127 n+128)
}{4838400}t^8-frac{(511 n+512) }{239500800}t^{10}+Oleft(t^{12}right)$$ 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

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