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Given that

$$\sum_{n=0}^{\infty}\sum_{j=0}^{m}(-1)^j{m\choose j}{1\over 4n+2j+1}=S\tag1$$ Where $m\ge1$

How can we show that $$S=2^{m-3}\color{blue}{\pi}-\sum_{k=0}^{m-2}{m-1\choose k+1}\sum_{l=0}^{k}{(-1)^l\over 2l+1}\color{red}?$$

Given examples:

$m=1:$

$$\sum_{n=0}^{\infty}\left({1\over 4n+1}-{1\over 4n+3}\right)={\pi\over 4}$$

$m=2:$

$$\sum_{n=0}^{\infty}\left({1\over 4n+1}-{2\over 4n+3}+{1\over 4n+5}\right)={\pi\over 2}-1$$

$m=3:$

$$\sum_{n=0}^{\infty}\left({1\over 4n+1}-{3\over 4n+3}+{3\over 4n+5}-{1\over 4n+7}\right)=\pi-{7\over 3}$$

Change $(1)$ into

$$\sum_{n=0}^{\infty}\sum_{j=0}^{m}(-1)^j{m\choose j}{1\over 4n+2j+1}=\sum_{j=0}^{m}(-1)^j{m\choose j}\int_{0}^{1}{x^{2j}\over 1+x^4}\tag2$$

2 Answers2

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We have the sum $S=\sum_{n=0}^{\infty} \sum_{j=0}^{m} \frac{(-1)^j \binom{m}{j}}{4n+2j+1}$. Now do the usual trick to deal with the denominator & sum the geometric & binomial series \begin{eqnarray*} \frac{1}{4n+2j+1}&=&\int_0^1 x^{4n+2j} dx \\ S &=& \int_0^1 dx \sum_{n=0}^{\infty} \sum_{j=0}^{m} (-1)^j \binom{m}{j} x^{4n+2j} \\ &=& \int_0^1 dx \frac{(1-x^2)^{m}}{(1-x^4)} = \int_0^1 dx \frac{(1-x^2)^{m-1}}{(1+x^2)} \end{eqnarray*} Now binomially expand the numerator & do an anti telescoping sum $x^{2k}=(-1)^k+\sum_{l=1}^{k}(-1)^{k-l}(x^{2l}+x^{2l-2})$ \begin{eqnarray*} S &=& \int_0^1 dx \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \left( \frac{(-1)^k}{1+x^2}+\sum_{l=1}^{k} (-1)^{k-l} x^{2l-2} \right) \\ &=& \sum_{k=0}^{m-1} \binom{m-1}{k} \left(\frac{\pi}{4}+\sum_{l=1}^{k} \frac{(-1)^l} {2l-1} \right) \\ &=& 2^{m-3}\pi+ \sum_{k=0}^{m-1} \binom{m-1}{k} \sum_{l=1}^{k} \frac{(-1)^l}{2l-1} \\ \end{eqnarray*}

Donald Splutterwit
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It appears there is an alternate closed form when $m$ is odd. Start by evaluating the inner sum:

$$\sum_{j=0}^m (-1)^j {m\choose j} \frac{1}{2j+4n+1}$$

using

$$f(z) = (-1)^m m! \frac{1}{2z+4n+1} \prod_{q=0}^m \frac{1}{z-q}.$$

We get for $0\le j\le m$

$$\mathrm{Res}_{z=j} f(z) = (-1)^m m! \frac{1}{2j+4n+1} \prod_{q=0}^{j-1} \frac{1}{j-q} \prod_{q=j+1}^m \frac{1}{j-q} \\ = (-1)^m m! \frac{1}{2j+4n+1} \frac{1}{j!} (-1)^{m-j} \frac{1}{(m-j)!} \\ = \frac{1}{2j+4n+1} (-1)^j {m\choose j}.$$

It follows that the inner sum is (residues sum to zero)

$$-\mathrm{Res}_{z=\infty} f(z) -\mathrm{Res}_{z=-2n-1/2} f(z).$$

We get for the first residue

$$-\mathrm{Res}_{z=\infty} f(z) = \mathrm{Res}_{z=0} \frac{1}{z^2} (-1)^m m! \frac{1}{2/z+4n+1} \prod_{q=0}^m \frac{1}{1/z-q} \\ = \mathrm{Res}_{z=0} (-1)^m m! \frac{1}{z^2} \frac{z}{2+z(4n+1)} \prod_{q=0}^m \frac{z}{1-qz} \\ = \mathrm{Res}_{z=0} (-1)^m m! \frac{1}{2+z(4n+1)} \prod_{q=1}^m \frac{z}{1-qz} \\ = \mathrm{Res}_{z=0} (-1)^m m! z^{m} \frac{1}{2+z(4n+1)} \prod_{q=1}^m \frac{1}{1-qz} = 0$$

because $m\ge 1.$ The second residue yields

$$- \frac{1}{2} (-1)^m m! \prod_{q=0}^m \frac{1}{-2n-1/2-q} = 2^m (-1)^{m+1} m! \prod_{q=0}^m \frac{1}{-4n-1-2q} \\ = 2^m m! \prod_{q=0}^m \frac{1}{4n+1+2q}.$$

and this is the value of the inner sum. Observe that

$$\prod_{q=0}^m \frac{1}{4(-n-(m+1)/2)+1+2q} = \prod_{q=0}^m \frac{1}{-4n+2(q-(m+1))+1} \\ = (-1)^{m+1} \prod_{q=0}^m \frac{1}{4n+2(m+1-q)-1} = (-1)^{m+1} \prod_{q=0}^m \frac{1}{4n+2(m-q)+1} \\ = \prod_{q=0}^m \frac{1}{4n+2q+1}.$$

It now follows that with

$$g(z) = 2^m m! \times \pi \cot(\pi z) \prod_{q=0}^m \frac{1}{4z+2q+1} \\ = \frac{1}{2^{m+2}} m! \times \pi \cot(\pi z) \prod_{q=0}^m \frac{1}{z+(2q+1)/4}$$

we have

$$2S + \sum_{p=1}^{(m-1)/2} \mathrm{Res}_{z=-p} g(z) + \sum_{p=0}^m \mathrm{Res}_{z=-p/2-1/4} g(z) = 0.$$

Note that

$$\mathrm{Res}_{z=-p/2-1/4} g(z) \\ = \frac{1}{2^{m+2}} m! \times \pi \cot(-\pi(2p+1)/4) \\ \times \prod_{q=0}^{p-1} \frac{1}{-(2p+1)/4+(2q+1)/4} \prod_{q=p+1}^{m} \frac{1}{-(2p+1)/4+(2q+1)/4} \\ = \frac{1}{2^{m+2}} m! \times \pi (-1)^{p+1} \prod_{q=0}^{p-1} \frac{2}{q-p} \prod_{q=p+1}^{m} \frac{2}{q-p} \\ = \frac{\pi}{4} m! \times (-1)^{p+1} \frac{(-1)^p}{p!}\frac{1}{(m-p)!} = - \frac{\pi}{4} {m\choose p}.$$

Hence

$$\sum_{p=0}^m \mathrm{Res}_{z=-p/2-1/4} g(z) = - \frac{\pi}{4} \sum_{p=0}^m {m\choose p} = - \frac{\pi}{4} 2^m.$$

and we may conclude that

$$\bbox[5px,border:2px solid #00A000]{ S = \frac{\pi}{8} 2^m -\frac{1}{2} 2^m m! \sum_{p=1}^{(m-1)/2} \prod_{q=0}^m \frac{1}{-4p+2q+1}, \quad m\;\text{odd.}}$$

The required bounds on $g(z)$ are examined at the following MSE link.

Marko Riedel
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