Question about series involving zeta function 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?

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! 
