I wonder if it is possible to evaluate explicitly the sum $$S(N):=\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}\left(1-\frac{2j}{N}\right)^{N+1},\quad N\in\mathbb{N}.$$ In the large $N$ limit the result is straightforward if one approximates the argument of the sum with the exponential and then evaluates the sum as a geometric one $$S(N)\approx\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}e^{-\frac{2(N+1)}{N}j}=\frac{1-e^{-N+\frac{1}{N}}}{e^{2+\frac{2}{N}}-1}\xrightarrow{N\to\infty}\frac{1}{e^2-1}.$$ In the general case I tried to manipulate the sum in some ways, considering separately odd or even $N$.
In the first case I wrote $S(N)$ in terms of the generalized Zeta function $$\zeta(s,a)=\sum_{k=0}^{\infty}(k+a)^{-s},$$ starting as follows $$\sum_{j=1}^{\frac{N-1}{2}}\left(1-\frac{2j}{N}\right)^{N+1}=\sum_{j=1}^{\frac{N-1}{2}}\left(\frac{2j-1}{N}\right)^{N+1}=\frac{1}{N^{N+1}}\left[\sum_{j=1}^{\infty}(2j-1)^{N+1}-\sum_{j=\frac{N+1}{2}}^{\infty}(2j-1)^{N+1}\right].$$ Note that the first passage holds because the summed terms are exactly the same, with only their order changed. The first term in the braket is $0$ for odd $N$ because it can be rewritten as $$\sum_{j=1}^{\infty}j^{N+1}-\sum_{j=1}^{\infty}(2j)^{N+1}=\left(1-2^{N+1}\right)\zeta(-N-1)$$ but $\zeta(-2m)=0,\;m\in\mathbb{N}$. The change of variable $j\rightarrow j-(N+1)/2$ on the remaining term leads to the result $$S(N)=-\left(\frac{2}{N}\right)^{N+1}\zeta\left(-N-1,\frac{N}{2}\right),\quad N\;\rm{odd}.$$
Another possibility is to use the generalized harmonic numbers $$H_n^{(m)}=\sum_{k=1}^n\frac{1}{k^m}.$$ Doing so for odd $N$ it holds $$S(N)=\frac{1}{N^{N+1}}\sum_{j=1}^{\frac{N-1}{2}}(2j-1)^{N+1}=\frac{1}{N^{N+1}}\left(\sum_{j=1}^{N-2}j^{N+1}-\sum_{j=1}^{\frac{N-3}{2}}(2j)^{N+1}\right)=\frac{1}{N^{N+1}}\left(H_{N-2}^{(-N-1)}-2^{N+1}H_{\frac{N-3}{2}}^{(-N-1)}\right).$$ Instead for even $N$ $$S(N)=\frac{1}{N^{N+1}}\sum_{j=1}^{\frac{N}{2}-1}(2j)^{N+1}=\left(\frac{2}{N}\right)^{N+1}H_{\frac{N}{2}-1}^{(-N-1)}.$$ I can't figure out if further manipulations are possible to simplify my results, or if a completely different approach may exist to treat such a sum.
The finite sum can be expressed in terms of Bernoulli polynomial $B_p(x)$. Bernoulli polynomials obey many interesting relations, the one we need is
$$B_p(x+1) - B_p(x) = px^{n-1}$$
Let $J = \lfloor \frac{N-1}{2}\rfloor$ and using this relation, we find
$$\begin{align}S(N) &=\sum_{j=1}^J\left(1-\frac{2j}{N}\right)^{N+1}\\ &= \left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left(\frac{N}{2}-j\right)^{N+1}\\ &= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left[B_{N+2}\left(\frac{N}{2}-j+1\right) - B_{N+2}\left(\frac{N}{2} - j\right)\right]\\ &= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1}\left[B_{N+2}\left(\frac{N}{2}\right) - B_{N+2}\left(\frac{N}{2}-J\right)\right] \end{align} $$