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After some calculations with WolframAlfa, it seems that

$$ \frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}} $$ Where $$ \eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}} $$ is the Dirichlet Eta function.

Could it be proved that this is true, or false?

Thanks.


ADDED:

If we consider the Dirichlet Beta function $$ \beta(z)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{z}} $$ We can write this as

$$ \beta(1)=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}} $$


ADDED:

I recently also noted that $$ \frac{\pi}{4}=\sum_{k=1}^{\infty}\frac{\eta(k)}{2^{k}} $$

So summing both of the expressions Whe have that $$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $$ and $$ \frac{1}{2}=\sum_{k=1}^{\infty}\frac{\eta(2k-1)}{2^{2k-1}} $$

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By changing the order of summation, you get the Leibniz series for $\frac{\pi}{4}$. Since we don't have absolute convergence, the change of order of summation must be justified. I expect it can be justified. –  Daniel Fischer Sep 3 '13 at 15:07
    
@DanielFischer can you post to show how Leibniz is obtained? It's almost a complete answer assuming the rearrangement can be justified (it would be crazy if it couldn't be, since the answer for the sums is the same). Maybe someone else can take your rearrangement strategy and fill in the proof for justifying rearrangement preserves the sum. –  user2566092 Sep 3 '13 at 15:09
    
The justification was surprisingly simple. –  Daniel Fischer Sep 3 '13 at 15:53
    
Yes (after seeing it), but I wonder if this is new... –  Neves Sep 3 '13 at 15:55
3  
I bet Euler knew it already ;) –  Daniel Fischer Sep 3 '13 at 15:57

2 Answers 2

up vote 34 down vote accepted

If we can change the order of summation, we obtain

$$\begin{align} 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1} \left(-\frac{1}{2n}\right)\frac{1}{1 + \frac{1}{2n}}\\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1}, \end{align}$$

which is the Leibniz series for $\frac{\pi}{4}$.

The convergence of the double sum is not absolute, so the change of summation requires a justification. We obtain that by a slightly more circumspect computation:

$$\begin{align} 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty \frac{(-1)^k}{2^k}\eta(k)\\ &= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty\frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k}\\ &= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=2}^\infty \frac{(-1)^k}{(2n)^k}\\ &= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\left(-\frac{1}{2n}\right)^2\frac{1}{1+\frac{1}{2n}}\\ &= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\ &= 1 - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{2n(2n+1)} -\frac{1}{2n}\right)\\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1}. \end{align}$$

Here the change of order of summation is unproblematic, since

$$\sum_{n=1}^\infty\sum_{k=2}^\infty \frac{1}{(2n)^k} = \sum_{n=1}^\infty \frac{1}{2n(2n-1)} < \infty.$$

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This question is an opportunity to showcase Mellin transforms and harmonic sums, where we first compute the Mellin transform of the sum and subsequently invert it, obtaining an asymptotic expansion about zero/infinity. Consider $$g(x) = \frac{1}{1+x}.$$ The Mellin transform $g^*(s)$ of $g(x)$ is given by $$g^*(s) = \mathfrak{M}(g(x); s) = \int_0^\infty \frac{1}{1+x} x^{s-1} dx.$$

Use a keyhole contour with the slot for the key on the postive real axis to evaluate this integral, where the branch of the logarithm for $x^{s-1} = e^{\log(x) (s-1)}$ has the cut along the positive real axis and produces arguments from $0$ to $2\pi$, getting $$ g^*(s) \left(1 - e^{2\pi i (s-1)}\right) = 2\pi i \operatorname{Res}\left( \frac{1}{1+x} x^{s-1} ; s=-1\right).$$ This implies $$ g^*(s) \left(1 - e^{2\pi i s}\right) = 2\pi i \times e^{\pi i (s-1)}.$$ Therefore we have $$g^*(s) = 2\pi i\frac{e^{-\pi i} e^{\pi i s}}{1 - e^{2\pi i s}} = -\pi \frac{2i e^{\pi i s}}{1 - e^{2\pi i s}} = -\pi \frac{2i}{e^{-\pi i s} - e^{\pi i s}} = \frac{\pi}{\sin(\pi s)}.$$

Now recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1}\lambda_k f(\mu_k x); s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}\right) f^*(s).$$ Put $$\lambda_k = (-1)^{k+1}, \quad \mu_k = k \quad\text{and}\quad f(x) = \frac{1}{1+2x}$$ so that $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \eta(s).$$ This yields $$\mathfrak{M}\left(\sum_{k\ge 1} (-1)^{k+1} \frac{1}{1+2kx}; s \right) = \frac{\eta(s)}{2^s} \frac{\pi}{\sin(\pi s)}.$$

The transform of $g(x)$ has fundamental strip $\langle 0,1\rangle$ and we may apply Mellin inversion to this transform to recover an expansion about infinity of the harmonic sum. This yields $$\sum_{k\ge 1} (-1)^{k+1} \frac{1}{1+2kx} = -\sum_{q\ge 1} \operatorname{Res}\left(\frac{\eta(s)}{2^s} \frac{\pi}{\sin(\pi s)}/x^s; s=q\right) = -\sum_{q\ge 1} \frac{\eta(q)}{2^q} \frac{(-1)^q}{x^q}.$$ It follows that $$1 + \sum_{q\ge 1} \frac{\eta(q)}{2^q} \frac{(-1)^q}{x^q} = 1 + \sum_{k\ge 1} (-1)^k \frac{1}{1+2kx} = \sum_{k\ge 0} (-1)^k \frac{1}{1+2kx}.$$ Finally put $x=1$ to obtain that $$ 1 + \sum_{q\ge 1} (-1)^q\frac{\eta(q)}{2^q} = \sum_{k\ge 0} (-1)^k \frac{1}{1+2k} = \frac{\pi}{4}.$$

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The forward transform can also be computed with the substitution $x = \frac{t}{t + 1}$, yielding $\int_0^{1}x^{s-1}(1-x)^{-s} = B(s, 1-s) = \Gamma(s)\Gamma(1-s)$. –  Bitrex Sep 4 '13 at 22:22

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