Simplify a double summation involving binomial coeficient

Question

$$T(N,K)=\sum_{i=2}^{K}\sum_{j=2}^{i}(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}$$ Is it possible to evaluate the sum for $K=10^7$ efficiently. If we manage to remove one of the sums, it will be doable.

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Substituting $i-j=u$ $$T(N,K)=\sum_{i=2}^{K}\sum_{j=2}^{i}(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}$$ $$T(N,K)=\sum_{i=2}^{K}\sum_{u=0}^{i-2}(-1)^u\binom{i}{u}\frac{(i-u)^{N+1}-1}{i-u-1}$$ $$T(N,K)=\sum_{i=2}^{K}\sum_{u=0}^{i-2}(-1)^u\binom{i}{u}\sum_{m=0}^{N}(i-u)^m$$ $$T(N,K)=\sum_{i=2}^{K}\sum_{m=0}^{N}\sum_{u=0}^{i-2}(-1)^u\binom{i}{u}(i-u)^m$$ $$T(N,K)=\sum_{i=2}^{K}\sum_{m=0}^{N}i!\left(\frac{1}{i!}\sum_{u=0}^{i-2}(-1)^u\binom{i}{u}(i-u)^m\right)$$ $$T(N,K)=\sum_{i=2}^{K}\sum_{m=0}^{N}i!\left(S_2(m,i)-\frac{(-1)^{i-1}}{(i-1)!}\right)$$ $S_2(m,i)$ denotes Stirling numbers of second kind. I can't proceed further. Any help is much appreciated.

Answer 1

one sum can be carried out: exchanging the order of summation, $$T(N,K)=\sum_{j=2}^K\sum_{i=j}^K(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}=$$ $$=\sum_{j=2}^K\frac{j^{N+1}-1}{j-1}\left[\frac{1}{2^{j+1}}+(-1)^{K-j} \binom{K+1}{j} \, _2F_1(1,K+2;K+2-j;-1)\right]$$