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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.

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    This question would be greatly improved by additional context. Please see "How to ask a good question" at http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question . You can edit the question to add information such as the source of the problem, the motivation behind it, and any attempts you have made to solve it. – 2017-01-15
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    Very similar to the [Euler sum](https://en.wikipedia.org/wiki/Euler_summation)... – 2017-01-15
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    Using mookid's notation, note that $$S_{1}+S_{2}=B_{n}$$ where $B_{n}$ are the Bernoulli numbers. – 2017-01-16
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    Well, They are from Bernoulli numbers. I want to used them separately. – 2017-01-16
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    @user90369 Uh? $S_{1}+S_{2}$ is the forumla $(33)$ here http://mathworld.wolfram.com/BernoulliNumber.html and however $S_{1}$ and $S_{2}$ are real numbers. How can the sum of two real number be a complex number with imaginary part $\neq 0$? – 2017-01-26
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    @MarcoCantarini : Thanks! Sorry, I have calculated $(-1)^{r/2}$ instead of $(-1)^r$ . Therefore $\displaystyle S_1=-\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is odd}}^k{k\choose r}r^n$ and $\displaystyle S_2=\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is even}}^k{k\choose r}r^n$ . Now it's easier to understand. :-) – 2017-01-26
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    @user90369 No problem, details are important in mathematics :) – 2017-01-26
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    I don't know if it is somehow related, but I had noticed that : $${\sum_{k=1}^n\frac{1}{k}\sum_{r=1}^k(-1)^r{k\choose r}r^n}=0$$ for all n $n\ge 2$ – 2017-01-26

2 Answers 2

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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 $$S_1 + i S_2$$?

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    Can you please answer your question ? :-) – 2017-01-26
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    @user90369 maybe answer their question? – 2017-01-26
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    @JpMcCarthy : I've got the impression that none of the answers here have something to do with the question of the OP in a aim-oriented manner. :-) – 2017-01-27
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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.

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    Which influence do have all your considerations for the term $\binom{k}{r}r^n$ in the original question ? :-) – 2017-01-27
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    @Basti Why you omit $r^n$ from $\sum_{k=0}^{n} \frac{1}{k+1}\sum_{r=0\\ r\ even}^{k} \binom{n}{k}(-1)^r$. – 2017-01-27
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    hahaha i just realized that i didnt see the $r^n$ factor. – 2017-01-27