Understanding a summation involving "$n$ choose $k$" that starts from $k=0$ and then $k$ goes negative?

Question

Algorithm 3 in Peter Borwein's paper (2000) offers this :- $$\zeta (s)=\Biggl(\frac{1}{2^n(1 - 2^{1 - s})}\sum_{j=0}^{2n-1}\frac{e_j}{(j+1)^s}\Biggr)+\gamma_{n}(s)$$ Where $$e_j=(-1)^j\Biggl(\sum_{k=0}^{j-n}\frac{n!}{k!(n-k)!}-2^n\Biggr)$$ The outer sum is varying $j$ over the range $0\le j\le(2n-1)$ while the inner sum is varying $k$ over the range $0\le k\le (j-n)$ but when $j \lt n$ the "upper" bound is negative ! $\;$ See example below for $n=6$ :- $\text {For} \; \displaystyle j=11\quad \sum_{k=0}^{5}\binom 6k-2^6 \quad = \: \text {(1+6+15+20+15+6)} \, -2^6$ $\text {For} \; \displaystyle j=9 \; \; \quad \sum_{k=0}^{3}\binom 6k-2^6 \quad = \: \text {(1+6+15+20)} \, -2^6$ $\text {For} \; \displaystyle j=6 \; \; \quad \sum_{k=0}^{0}\binom 6k-2^6 \quad = \: \text {(1)} \, -2^6$ $\text {When} \; j \lt 6 \; $ should I disregard the summation and interpret $ \; e_j \; \text {as} \; (-1)^j(-2^n) \; \text {?}$