Is $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{n^2+1}$ related to any popular constants Wolfram-Alpha
shows this to be converging to roughly 0.2696:$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{n^2+1}$$
Is this related to other known constants?
 A: I suppose I can't let everyone post answers with no explanations now can I?
First, apply PFD to get
$$\frac n{n^2+1}=\frac12\left(\frac1{n+i}+\frac1{n-i}\right)$$
Thus, the given sum is equivalent to
$$S=\frac12\sum_{n=0}^\infty\left(\frac{(-1)^{n+1}}{n+i}+\frac{(-1)^{n+1}}{n-i}\right)$$
It is then straightforward to note that:
$$\Phi(z,s,\nu)=\sum_{n=0}^\infty\frac{z^n}{(n+\nu)^s}$$
Where $\Phi$ is the Lerch Transcendent.  Thus, one form of your sum is
$$S=-\frac{\Phi(-1,1,i)+\Phi(-1,1,-i)}2$$
Another form follows from the digamma function.  To see this one, we split the sum over multiple parts:
$$2S=\sum_{n=1}^\infty\left(\frac1{2n+i-1}-\frac1{2n+i}+\frac1{2n-i-1}-\frac1{2n-i}\right)$$
$$4S=\sum_{n=1}^\infty\left(\frac1{n+\frac{i-1}2}-\frac1{n+\frac i2}+\frac1{n-\frac{i+1}2}-\frac1{n-\frac i2}\right)$$
Now we exploit the series expansion of the digamma function:
$$\psi(z+1)=-\gamma+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right)$$
Thus, we obtain Mathematica's result:
$$4S=\psi\left(1+\frac i2\right)-\psi\left(\frac{1+i}2\right)+\psi\left(1-\frac i2\right)-\psi\left(\frac{1-i}2\right)$$
By exploiting the recurrence formula
$$\psi(z+1)=\psi(z)+\frac1z$$
and the reflection formula
$$\psi(1-z)-\psi(z)=\pi\cot(\pi z)$$
We find that, equivalent to Robert Israel's answer,
$$4S=2i+2\psi\left(1+\frac i2\right)-\pi i\coth\left(\frac\pi2\right)-2\psi\left(\frac{1+i}2\right)+\pi i\tanh\left(\frac\pi2\right)$$
Since we know the sum is real,
$$S=\frac12\Re\left[\psi\left(1+\frac i2\right)-\psi\left(\frac{1+i}2\right)\right]$$
A: It is the real part of 
$$\frac{1}{2} \left(\Psi\left(\frac{i}2\right) -\Psi\left(\frac{1+i}{2}\right)\right) = \Psi\left(\frac{i}2\right) - \Psi(i) + \ln(2)$$ 
It could also be written as
$$ - \frac{\Phi(-1,1,1-i)+\Phi(-1,1,1+i)}{2}$$
where $\Phi$ is the Lerch Phi function.
A: According to Mathematica it is equal to
$$
\frac{1}{4} \left(-\psi ^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)-\psi
   ^{(0)}\left(\frac{1}{2}+\frac{i}{2}\right)+\psi ^{(0)}\left(1-\frac{i}{2}\right)+\psi
   ^{(0)}\left(1+\frac{i}{2}\right)\right),
$$
where $\psi$ denotes the polygamma function.
Numerically it is equal to 0.26961050270800898180..., which does not apear in the Inverse Symbolic Calculator , so it is not likely to be some 'known' number.
