Intereting Posts

Is the derivative of differentiable function $f:\mathbb{R}\to\mathbb{R}$ measurable on $\mathbb{R}$?
Integral formulas involving continued fractions
Known exact values of the $\operatorname{Li}_3$ function
Prove the function is not differentiable at (0,0)
Different definitions of trigonometric functions
Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
How to prove that $\cos^2(z)+\sin^2(z)=1$, where $z$ is a complex variable (if it is true)?
Is it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?
Density of Diagonalizable matrices
Diophantine approximation – Closest lattice point to a line (2d)
A question on definite integral to find a value
Formal Proof that area of a rectangle is $ab$
A number system
Given relations on matrices $H,V,$ and some vectors, can we deduce that $x = 0$?
Any example of a function which is discontinuous at each point in a deleted neighborhood of a point at which that function is differentiable?

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?

- Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$
- I can't remember a fallacious proof involving integrals and trigonometric identities.
- What are the uses of Euler's number $e$?
- How is $e$ the only number $n$ for which $n^x > x^n$ is satisfied for all values of $x$?
- Are my calculations of a recursive prime-generating function based on logarithms correct?
- Difference between variables, parameters and constants

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}}

$$

- Another simple rule satisfied by the Fibonacci $n$-step constants?
- Divergence of Dirichlet series
- How is $e$ the only number $n$ for which $n^x > x^n$ is satisfied for all values of $x$?
- Intuitive Understanding of the constant “$e$”
- What is circumradius $R$ of the great disnub dirhombidodecahedron, or Skilling's figure?
- Translations AND dilations of infinite series
- Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$
- Computing the Dirichlet Density
- Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$
- Difference between variables, parameters and constants

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}.$$

- How to determine if two points lie on a vector, given a unit vector
- Algebra: Best mental images
- Picturing the discrete metric?
- If $f(xy)=f(x)f(y)$, then $f(x)=x^b$.
- Show that $\arctan(n)$ is irrational for all $n \in \mathbb{N}$
- Explanation of proof of Gödel's Second Incompleteness Theorem
- An example of a Banach space isomorphic but not isometric to a dual Banach space
- Continuous and bounded imply uniform continuity?
- Adjoint of the covariant derivative on a Riemannian manifold
- $f(x-y)$ considered as a function of $(x,y)\in \mathbb{R^{2n}}$ is measurable if $f$ is measurable
- Cancellation in topological product
- How to prove triangle inequality for $p$-norm?
- A complex algebraic variety which is connected in the usual topology
- Convergence of the series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$; I say it converges, WebWork tells me this is incorrect.
- Solutions of $\arctan x = 1 – x$. Proofs?