Closed form for sum of powers 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
$$\begin{align}S(N)
&=\sum_{j=1}^J\left(1-\frac{2j}{N}\right)^{N+1}\\
&= \left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left(\frac{N}{2}-j\right)^{N+1}\\
&= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left[B_{N+2}\left(\frac{N}{2}-j+1\right) - B_{N+2}\left(\frac{N}{2} - j\right)\right]\\
&= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1}\left[B_{N+2}\left(\frac{N}{2}\right) - B_{N+2}\left(\frac{N}{2}-J\right)\right]
\end{align}
$$
