I was observing the series $$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)-1}{2n+1}$$ And Wolfram alpha says that it does not converge. But I'm convinced that this is wrong since $$\sum_{n=1}^{\infty} (\zeta(2n+1)-1) = \frac{1}{4}$$
I would imagine then, by comparison test, that the first series clearly converges. Am I wrong? If it does converge, do we have a value for it?
On
Hint. The convergence may be obtained by the comparison test using, as $n \to \infty$, $$ \frac{\zeta(2n+1)-1}{2n+1}\sim \frac{\frac1{2^{2n+1}}}{2n+1} $$ a closed form of the sum may be obtained by integrating the standard series $$ \psi(1+z)+\psi(1-z)= -2\gamma -2\sum_{k=1}^\infty \zeta (2k+1) z^{2k}, \qquad |z|<1, $$ where $\psi$ denotes the digamma function.
On
Perhaps I can fill in the missing details for you to help show how $$I = \int^\infty_0 \frac{\sinh x - x}{x} \cdot \frac{dx}{e^x (e^x - 1)} = 1 - \gamma - \frac{1}{2} \ln (2).$$
To do this we will follow Jack's suggestion and use Frullani's theorem together with the following integral representation for the Euler-Mascheroni constant of $$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx.$$
We begin by observering that $$\frac{1}{e^x (e^x - 1)} = \frac{1}{e^x - 1} - \frac{1}{e^x}.$$ So the integral can be rewritten as \begin{align*} I &= \int^\infty_0 \frac{\sinh x - x}{x} \left (\frac{1}{e^x - 1} - \frac{1}{e^x} \right ) \, dx\\ &= \int^\infty_0 \left [\frac{\sinh x - x}{x (e^x - 1)} - \frac{\sinh x - x}{x e^x} \right ] \, dx\\ &= \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{1}{e^x - 1} - \frac{\sinh x}{x e^x} + \frac{1}{e^x} \right ] \, dx\\ &= \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{1}{e^x - 1} - \frac{\sinh x + 1}{x e^x} + \frac{1}{xe^x} + e^{-x} \right ] \, dx\\ &= -\int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx + \int^\infty_0 e^{-x} \, dx + \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{\sinh x + 1}{x e^x} \right ] \, dx\\ &= -\gamma + 1 + \int^\infty_0 \frac{1}{x} \left [\frac{\sinh x}{e^x - 1} - \frac{\sinh x + 1}{e^x} \right ] \, dx\\ &= 1 - \gamma + I_\alpha. \end{align*}
Now for the term appearing in the square brackets of the integral, it can be rewritten as \begin{align*} \frac{\sinh x}{e^x - 1} - \frac{\sinh x + 1}{e^x} &= \frac{e^x \sinh x - (e^x - 1)(\sinh x + 1)}{e^x (e^x - 1)}\\ &= \frac{\sinh x - e^x + 1}{e^x (e^x - 1)}\\ &= \frac{\frac{1}{2} (e^x - e^{-x}) - e^x + 1}{e^x (e^x - 1)}\\ &= -\frac{e^{-2x} - 2e^{-x} + 1}{2(e^x - 1)}\\ &= -\frac{(1 - e^{-x})^2}{2 e^x (1 - e^{-x})}\\ &= \frac{e^{-2x} - e^{-x}}{2}. \end{align*}
Now as $$I_\alpha = \int^\infty_0 \frac{e^{-2x} - e^{-x}}{x} \, dx,$$ is of the form of a Frullani integral, namely $$\int^\infty_0 \frac{f(ax) - f(bx)}{x} \, dx = (f(0) - f(\infty)) \ln \left (\frac{b}{a} \right ),$$ where $f(x) = e^{-x}, a = 2, b = 1$, as $f(0) = 1$ and $f(\infty) = 0$ we have $$I_\alpha = \ln \left (\frac{1}{2} \right ) = -\ln (2).$$ Thus $$\int^\infty_0 \frac{\sinh x - x}{x} \cdot \frac{dx}{e^x (e^x - 1)} = 1 - \gamma - \frac{1}{2} \ln (2),$$ and is all thanks to Jack and his amazing insight!
As you noticed, since the series $\sum_{n\geq 1}\left[\zeta(2n+1)-1\right]$ has bounded partial sums, the series $\sum_{n\geq 1}\frac{\zeta(2n+1)-1}{2n+1}$ is convergent by Dirichlet's test.
By the integral representation for the $\zeta(s)$ function in the region $s>1$ (i.e. by the inverse Laplace transform) we have $$ \zeta(2k+1) =\int_{0}^{+\infty}\frac{x^{2k}}{(2k)!}\cdot \frac{dx}{e^x-1} $$ $$ \zeta(2n+1)-1 =\int_{0}^{+\infty}\frac{x^{2n}}{(2n)!}\cdot \frac{dx}{e^x(e^x-1)} \tag{A}$$ $$ \frac{\zeta(2n+1)-1}{2n+1} =\int_{0}^{+\infty}\frac{x^{2n}}{(2n+1)!}\cdot \frac{dx}{e^x(e^x-1)} \tag{B}$$ $$ \sum_{n\geq 1}\frac{\zeta(2n+1)-1}{2n+1} =\int_{0}^{+\infty}\frac{\sinh(x)-x}{x}\cdot \frac{dx}{e^x(e^x-1)} \tag{C}$$ and the RHS of $(C)$ equals $\color{red}{1-\gamma-\frac{1}{2}\log 2}\approx 0.07621$ by Frullani's theorem and the integral representation for the Euler-Mascheroni constant. Definitely a convergent series, and a curious bug of WA.