What's a closed form for $\sum_{k=0}^n\frac{1}{k+1}\sum_{r=0\\r~is~odd}^k(-1)^r{k\choose r}r^n$

Question

I want to use a closed form of $$\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is odd}}^k(-1)^r{k\choose r}r^n$$ and $$\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is even}}^k(-1)^r{k\choose r}r^n$$ Thanks.

Answer 1

Hint: $$S_1 = \sum_{k=0}^n\frac{1}{k+1}\sum_{r=0\\r~is~odd}^k(-1)^r{k\choose r}r^n$$

$$S_2 = \sum_{k=0}^n\frac{1}{k+1}\sum_{r=0\\r~is~even}^k(-1)^r{k\choose r}r^n$$

What is $$S_1 + i S_2$$?

Answer 2

By the binomial formula, we have: $$ \sum_{k=0}^{n} \binom{n}{k} = \sum_{k=0}^{n} \binom{n}{k}1^{n-k}1^{k} = (1+1)^n=2^n$$ In a similar way: $$ \sum_{k=0}^{n} \binom{n}{k}(-1)^k = (1-1)^n = 0$$ Adding and subtracting these formulas, we deduce $$ \sum_{k=0\\ k\ odd}^{n} \binom{n}{k} = \sum_{k=0\\ k\ even}^{n} \binom{n}{k} = 2^{n-1}$$ Since $(-1)^r=1$ when $r$ is even, it follows: $$\sum_{k=0}^{n} \frac{1}{k+1}\sum_{r=0\\ r\ even}^{k} \binom{n}{k}(-1)^r = \sum_{k=0}^{n} \frac{2^{k-1}}{k+1}=\frac{1}{4}\sum_{k=0}^{n} \frac{2^{k+1}}{k+1}=\frac{1}{4}\int_{0}^{2}\left ( \sum_{k=0}^{n} x^{k} \right )=\\ \frac{1}{4}\int_{0}^{2} \frac{x^{n+1}-1}{x-1}$$

pd. I think you was asking for $$ \sum_{k=0}^{n}\frac{1}{k+1}\left (\binom{k}{0}-\binom{k}{2}+\binom{k}{4}-...\right )$$ In that case, thinking in $S_{1}+iS_{2}$ works. One obtain something like $$ S_{1}+iS_{2} = \sum_{k=0}^{n}\sum_{r=0}^{k} \frac{e^{\frac{i\pi r}{2}}}{k+1}$$ It can be manipulated by the same techniques showed here.