Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots After some calculations with WolframAlfa, it seems that$$ \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}} $$Solutions Collecting From Web of "Proving that \frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$"

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

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

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
$$1 + \sum_{q\ge 1} (-1)^q\frac{\eta(q)}{2^q} = \sum_{k\ge 0} (-1)^k \frac{1}{1+2k} = \frac{\pi}{4}.$$