If $\displaystyle S=\sum^{n}_{k=0}(-1)^k\frac{1}{k+m+1}\binom{n}{k}$ and f $\displaystyle T=\sum^{m}_{k=0}(-1)^k\frac{1}{k+n+1}\binom{m}{k}$.Then $S-T$ is
Try:$$S=\sum^{n}_{k=0}(-1)^k\int^{1}_{0}x^{k+n}\binom{n}{k}dx=\int^{1}_{0}x^n\sum^{n}_{k=0}(-x)^k\binom{n}{k}dx$$
Could some help me, thanks
$\def\d{\mathrm{d}}$\begin{align*} S &= \sum_{k = 0}^n (-1)^k \frac{1}{k + m + 1} \binom{n}{k} = \sum_{k = 0}^n (-1)^k \binom{n}{k} \int_0^1 x^{k + m} \,\d x\\ &= \int_0^1 \sum_{k = 0}^n (-1)^k \binom{n}{k} x^{k + m} \,\d x = \int_0^1 x^m \sum_{k = 0}^n \binom{n}{k} (-x)^k \,\d x\\ &= \int_0^1 x^m (1 - x)^n \,\d x, \end{align*} and analogously,$$ T = \int_0^1 x^n (1 - x)^m \,\d x. $$ Make substitution $u = 1 - x$ to get$$ S = \int_0^1 x^m (1 - x)^n \,\d x = \int_0^1 (1 - u)^m u^n \,\d u = \int_0^1 x^n (1 - x)^m \,\d x = T. $$
