closed form of $\sum_{k=1}^\infty\frac{(-1)^k}{2k-1}\ln(2k+1+n)$

Question

In my work on another problem I arrived at this sum:

$$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1}\ln(2k+1+n)$$

It appears to converge for practically any $n$. Is this a known sum that I can find references to? Does it have a closed form? Should I try converting the log into a series?

Answer 1

By the alternating series test, it converges for all $n>-3$ (and in fact for all $n\in \mathbb{C}$ such that all the terms of the series exist). A closed form is known for $n=-2$:$$ \sum_{k=1}^\infty \frac{(-1)^k}{2k-1} \log(2k-1) =\beta'(1)= \frac\pi4(\gamma - \log \pi) + \pi \log\Gamma(3/4) $$ according to Wikipedia.

I don't expect there's a nice closed form for any other special value of the sum. Let $f(z) = \sum_{k=1}^\infty \frac{(-1)^k}{2k-1}\log(2k-1+z)$; then your sum is $f(n+2)$ (it's slightly nicer to shift the argument). The closest thing to a closed form I could find for $f$ is it's Taylor series $$ f(z) = \beta'(1) + \sum_{n=1}^\infty \frac{\beta(n+1)}{n} (-z)^n $$ where $\beta$ is the Dirichlet Beta function. To prove this, recall $\log(a + z) = \log(a) - \sum_{n=1}^\infty \frac1{n a^n}(-z)^n$. Applying this to $f$:\begin{eqnarray} f(z) &=& \sum_{k=1}^\infty \frac{(-1)^k}{2k-1}\log(2k-1+z)\\ &=& \sum_{k=1}^\infty \frac{(-1)^k}{2k-1}\log(2k-1) - \sum_{k=1}^\infty \frac{(-1)^k}{2k-1} \sum_{n=1}^\infty \frac{1}{n(2k-1)^n} (-z)^n\\ &=& -\beta'(1) + \sum_{n=1}^\infty \frac{\beta(n+1)}{n}(-z)^n \end{eqnarray} Also closed form adjacent is $$ f'(z) = -\frac{\pi}{4z} - \frac{1}{4z}\left(\psi\left(\frac{z+1}4\right) - \psi\left(\frac{z+3}4\right)\right) $$ where $\psi$ is the digamma function. Proof:\begin{eqnarray} f'(z) &=& \sum_{k=1}^\infty \frac{(-1)^k}{(2k-1)(2k-1+z)}\\ &=& \sum_{k=1}^\infty (-1)^k\left(\frac1{z(2k-1)} - \frac1{z(2k-1+z)} \right) \\ &=& \frac1z\sum_{k=1}^\infty \frac{(-1)^k}{2k-1} - \frac1z\sum_{k=1}^\infty \frac{1}{2k-1+z} \\ &=& -\frac{\pi}{4z}- \frac{1}{4z}\left(\psi\left(\frac{z+1}4\right) - \psi\left(\frac{z+3}4\right)\right) \end{eqnarray} see Wolfram Alpha for the last step.

Answer 2

This isn't an answer, but a possible numerical strategy.

Given the following numerical series:

$$ \sum_{k = 1}^{\infty} (-1)^k\,\frac{\ln(2\,k + 1 + n)}{2\,k - 1} $$

with $n > -3$ fixed, being a series with alternate sign terms, it can be calculated numerically using an efficient algorithm: the method of Henri Cohen, Fernando Rodriguez Villegas and Don Zagier.

In particular, for $n = 17$, writing in Wolfram Mathematica 12.0:

CVZ[fct_, {var_, varmin_}, niter_] :=
    Module[{a, b, c, d, n},
            a = ((3 + Sqrt[8])^niter + (3 - Sqrt[8])^niter) / 2;
            b = -1;
            c = -a;
            d = 0;
            Table[c = b - c;
                  d = d + c fct (-1)^varmin /. {var -> n + varmin};
                  b = (n + niter) (n - niter) b / ((n + 1/2) (n + 1)),
                 {n, 0, niter - 1}];
            N[{d/a, Abs[d/a^2]}]
          ];

CVZ[Log[2 k + 1 + 17] / (2 k - 1), {k, 1}, 8]

you get immediately:

{-2.33885, 3.51254*10^-6}

which are the sum and the respective maximum error made with only $8$ iterations.

Also, through this other simple code:

data = Table[CVZ[Log[2 k + 1 + n] / (2 k - 1), {k, 1}, 8], {n, -2, 1000}];

ListPlot[data[[All, 1]], PlotRange -> All]

ListPlot[data[[All, 2]], PlotRange -> All]

you get immediately:

where the $8$ iterations remained fixed; by increasing the number of iterations the maximum errors committed will decrease, while the calculation times will increase.