# 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?

• It is indeed a convergent sequence. Why would you expect that it is related to any known constant? – Jack Jun 30 '17 at 21:41
• I doubt it. Mathematica expresses this sum in terms of the digamma function. – José Carlos Santos Jun 30 '17 at 21:43
• @Jack Sometimes WA gives longer expressions for something simple because of the way the query is framed. I just wanted to rule out that. – Srini Jun 30 '17 at 21:50
• Its pretty close to the the formula for $\pi/4$, aka the alternating harmonic series. – Jonathan Davidson Jun 30 '17 at 21:52
• Note $\sum_{n=1}^\infty \frac{(-1)^{n} (2n-1)}{(2n-1)^2+1} = \frac12 \sum_{n=-\infty}^\infty \frac{(-1)^{n} (2n-1)}{(2n-1)^2+1}$ can be expressed in term of trigonometric functions – reuns Jun 30 '17 at 22:30

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,

$$S=\frac12\Re\left[\psi\left(1+\frac i2\right)-\psi\left(\frac{1+i}2\right)\right]$$

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.

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.