Summation of binomial-like terms

Question

Simplification of the two expressions $$S_1=n\sum_{k=0}^{n-1} {n-1 \choose k} \frac{(-1)^k}{k+1}$$ and $$S_2=n\sum_{k=0}^{n-1} {n-1 \choose k} \frac{(-1)^k}{k+3}$$ It seems that $S_1=1$, for example; are these familiar forms, available in a list of summations?

Answer 1

Hint:

$$\binom{n-1}k\cdot\dfrac1{k+3}=\dfrac{(k+1)(k+2)}{n(n+1)(n+2)}\binom{n+2}{k+3}$$

Let $(k+1)(k+2)=(k+3)(k+2)+A(k+3)+B$

$k=-3\implies(-2)(-1)=B\iff B=2$

$k=-2\implies A(-2+3)+B=A+2\iff A=-4$

$$\implies(k+1)(k+2)\binom{n+2}{k+3}$$

$$=(k+3)(k+2)\binom{n+2}{k+3}+(-4)(k+3)\binom{n+2}{k+3}+2\binom{n+2}{k+3}$$

$$=(n+2)(n+1)\binom n{k+1}-4(n+2)\binom{n+1}{k+2}+2\binom{n+2}{k+3}$$

Now $\displaystyle\sum_{k=0}^{n-1}\binom{n+2}{k+3}(-1)^k$

$=\displaystyle\sum_{r=3}^{n+2}\binom{n+2}r(-1)^{r-3}$

$=-\displaystyle\sum_{r=3}^{n+2}\binom{n+2}r(-1)^r$

$\displaystyle=\sum_{r=0}^2\binom{n+2}r(-1)^r-(1-1)^{n+2}$

Can you take it from here?

Answer 2

For (a), note that $$ \frac{(-1)^{k}}{k+1} = -\int_{0}^{-1}x^k dx $$ You can interchange the order of summation and integration. Comute the sum to get the expression of the form $(1+x)^n$, then integrate.