# Two Euler sums each containing the reciprocal of the central binomial coefficient

Is it possible to find closed-form expressions for the following two Euler sums containing the reciprocal of the central binomial coefficient? $$1. \sum_{n = 0}^\infty \frac{(-1)^n H_n}{(2n + 1) \binom{2n}{n}} \qquad \text{and} \qquad 2. \sum_{n = 0}^\infty \frac{(-1)^n H_{2n + 1}}{(2n + 1) \binom{2n}{n}}$$ Here $$H_n$$ is the $$n$$th harmonic number.

The reason I am interested in these two sums is as follows. It arose while considering alternative ways to evaluate the integral given here $$\int_0^1 \frac{\ln x}{x^2 - x - 1} \, dx = \frac{\pi^2}{5 \sqrt{5}}.$$

One way, very similar to the answer given, is to use the dilogarithm machinery and is relatively simple.

A second way is as follows: \begin{align} \int_0^1 \frac{\ln x}{x^2 - x - 1} \, dx &= - \int_0^1 \frac{\ln x}{1 + (x - x^2)} \, dx\\ &= - \int_0^1 \ln x \sum_{n = 0}^\infty (-1)^n (x - x^2)^n \, dx \tag1 \\ &= - \sum_{n = 0}^\infty (-1)^n \int_0^1 \ln x \, x^n (1 - x)^n \, dx\tag2. \end{align} In (1) a geometric expansion has been used and is valid since $$|x - x^2| < 1$$ for $$0 < x < 1$$. In (2) the summation and integration has been interchanged and is valid due to the dominated convergence theorem.

Consider $$J(a) = \int_0^1 x^a (1 - x)^n \, dx, \quad a > 0.$$ Differentiating with respect to $$a$$ we have $$J'(a) = \int_0^1 \ln x \, x^a (1 - x)^n \, dx,$$ and we observe that $$J'(n) = \int_0^1 \ln x \, x^n (1 - x)^n \, dx.$$

Now $$J(a) = \operatorname{B} (a + 1, n + 1) = \frac{\Gamma (n + 1) \Gamma (a + 1)}{\Gamma (a + n + 2)}.$$ Thus $$J' (a) = \frac{\Gamma (n + 1) \Gamma (a + 1)}{\Gamma (a + n + 2)} \left [\psi (a + 1) - \psi (a + n + 2) \right ].$$ Here $$\psi (z)$$ is the digamma function. Setting $$a = n$$ then gives $$J'(n) = \frac{1}{(2n + 1) \binom{2n}{n}} \left [\psi (n + 1) - \psi (2n + 2) \right ] = \frac{1}{(2n + 1) \binom{2n}{n}} (H_n - H_{2n + 1}),$$ where we have used the result: $$\psi(x) = H_{n - 1} - \gamma$$, with $$\gamma$$ corresponding the the Euler–Mascheroni constant.

So on returning to our integral we see that $$\int_0^1 \frac{\ln x}{x^2 - x - 1} \, dx = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1) \binom{2n}{n}} (H_{2n + 1} - H_n) = \frac{\pi^2}{5 \sqrt{5}},$$ and brings me to the two Euler sums.

Thoughts on finding the sums

One thought is to somehow massage the result $$\sum_{n = 0}^\infty \frac{x^{2n + 2}}{(n + 1)(2n + 1) \binom{2n}{n}} = 4 \arcsin^2 \left (\frac{x}{2} \right ),$$ in conjunction with perhaps the result $$H_n = - n \int_0^1 x^{n - 1} \ln (1 - x) \, dx,$$ into a suitable form but the presence of the alternating term $$(-1)^n$$ is proving difficult.

• It may or may not be of much use, but $$H_n=\int_0^1\frac{x^{n}-1}{x-1}dx$$ – clathratus Feb 9 at 5:40
• I'm not sure why you're calling these "Euler sums". They don't look like anything at mathworld.wolfram.com/EulerSum.html Maybe "Apery-like sums" would be closer to standard practice, as in math.arizona.edu/~rta/001/sherman.travis/series.pdf – Gerry Myerson Feb 9 at 6:17
• @Gerry Myerson I think of an Euler sum as an infinite sum that involves the harmonic number $H_n$ or its generalisation $H^{(k)}_n$. – omegadot Feb 9 at 8:03
• Maybe this could be of help $(13.4.2)$. At least the sum you denoted as $1.$ appeared there even though it is not explicitely evaluated but used in order to evalutate a slightly different sum. – mrtaurho Feb 9 at 9:53

EDIT 12.02.2019

Inspired by the comment of omegadot "Based on the numerical evidence it seems correct with the final two expressions being far more complicated than I could have ever imagined. I am guessing it would also be an almighty herculean task to finally show $$p_2-p_1 = \frac{\pi^2}{5\sqrt{5}}$$"

I have solved both problems (without being Hercules).

First I showed, starting from my expressions $$(2c)$$ and $$(3c)$$, that indeed

$$p_2-p_1 = \frac{\pi^2}{5\sqrt{5}}\tag{e1}$$

Then simplifying $$p_1$$ using the identity mentioned in my comment

$$\text{Li}_2\left(\frac{1}{z}\right)+\text{Li}_2(z)+\frac{1}{2} \log ^2(-z)+\zeta (2)=0, z\lt 0\tag{e2}$$

and using $$(e1)$$ for $$p_2$$ I find with

$$q=\frac{1}{2} \left(\sqrt{5}-1\right)=\frac{1}{\phi}\tag{e3}$$

these simplified expressions (using $$q$$ or the golden ratio $$\phi$$, alternatively)

$$p_1 = \frac{4}{\sqrt{5}} \;\text{Li}_2\left(-q^2\right)+\frac{4 \log (q) \log (5 q)}{\sqrt{5}}+\frac{\pi ^2}{3 \sqrt{5}}\tag{2e}$$

$$p_1 = -\frac{4 \text{Li}_2\left(-\phi ^2\right)}{\sqrt{5}}-\frac{4 \log (\phi ) \log (5 \phi )}{\sqrt{5}}-\frac{\pi ^2}{3 \sqrt{5}}\tag{2f}$$

$$p_2 = \frac{4}{\sqrt{5}} \;\text{Li}_2\left(-q^2\right)+\frac{4 \log (q) \log (5 q)}{\sqrt{5}}+\frac{8 \pi ^2}{15 \sqrt{5}}\tag{3e}$$

$$p_2 = -\frac{4 \text{Li}_2\left(-\phi ^2\right)}{\sqrt{5}}-\frac{4 \log (\phi ) \log (5 \phi )}{\sqrt{5}}-\frac{2 \pi ^2}{15 \sqrt{5}}\tag{3f}$$

Remark

Observing that the parameter $$q$$ solves the equation $$x^2+x-1=0$$ led me to study the integral

$$i_2= \int_0^1 \frac{\log(x)}{x^2+x-1}\,dx\tag{e4}$$

which is similar to the integral

$$i_1= \int_0^1 \frac{\log(x)}{x^2-x-1}\,dx$$

which was the starting point in the post of omegadot.

I solved the (divergent) integral $$i_2$$ with a limiting procedure starting from a general polynomial of 2nd degree in the denominator which gives the result

$$i_2 = \frac{3 \pi ^2}{10 \sqrt{5}}+\frac{i \pi \log (q)}{\sqrt{5}}\tag{e5}$$

The real part is the principal value of $$i_2$$.

Note that this expression resembles the constant terms in $$p_{1,2}$$.

Original post, 11.02.2019

The question was "Is it possible to find closed-form expressions for the following two Euler sums containing the reciprocal of the central binomial coefficient?"

The answer is yes, and we provide both closed expressions.

These are given in both cases in terms of combinations of $$\log(.)$$ and the polylog function $$\text{Li}_2(.)$$. The arguments are expressions of the type $$a + b \sqrt{5}$$ with rational $$a$$ and $$b$$ where $$\sqrt{5}$$ can also be expressed in terms of the golden ratio.

Definitions

$$p_1= \sum_{n = 0}^\infty \frac{(-1)^n H_n}{(2n + 1) \binom{2n}{n}} \qquad \text{and} \qquad p_2= \sum_{n = 0}^\infty \frac{(-1)^n H_{2n + 1}}{(2n + 1) \binom{2n}{n}}$$

Here $$H_n$$ is the $$n$$th harmonic number.

Results

Using

$$H_{n}=\int_0^1 \frac{1-x^n}{1-x} \, dx\tag{1}$$

I find

For $$p_1$$

The sum under the integral is

$$s_1 = \frac{4 \left(\sqrt{5} \sqrt{x} \sqrt{x+4}\; \text{arcsinh}\left(\frac{1}{2}\right)-5\; \text{arcsinh}\left(\frac{\sqrt{x}}{2}\right)\right)}{5 (1-x) \sqrt{x} \sqrt{x+4}}\tag{2a}$$

We have to evaluate the integral

$$p_1 = \int_0^1 s_1 \,dx \tag{2b}$$

This can also be done in Mathematica to give

$$p_1 =-\frac{2}{\sqrt{5}} \left(\text{Li}_2\left(\frac{1}{2} \left(-3-\sqrt{5}\right)\right)-\text{Li}_2\left(\frac{1}{2} \left(-3+\sqrt{5}\right)\right)\tag{2c}\\+\log (25) \log \left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\right)$$

Numerically,

$$N(p_1) \simeq -0.12711517793300195$$

Notice that $$p_1$$ can also be expressed by the "golden ratio

$$\phi=\frac{1}{2}(1+\sqrt{5}) \simeq 1.618033988749895$$

as

$$p_1=-\frac{2}{\sqrt{5}} (-\text{Li}_2(\phi -2)+\text{Li}_2(-\phi -1)+4 \log (2 \phi -1) \log (\phi ))\tag{2d}$$

For $$p_2$$

the sum can be done

$$s_2 = \frac{1}{1-x} \sum_{n = 0}^\infty (-1)^n (1-x^{2n + 1})\frac{1}{(2n + 1) \binom{2n}{n}}\\= \frac{4 \left(\sqrt{5} \sqrt{x^2+4} \;\text{arcsinh}\left(\frac{1}{2}\right)-5 \; \text{arcsinh}\left(\frac{x}{2}\right)\right)}{5 (1-x) \sqrt{x^2+4}}\tag{3a}$$

We have to evaluate the integral.

$$p_2 = \int_0^1 s_2 \,dx\tag{3b}$$

Mathematica 8.0 can do the integral (version 10.1 can't, so it is advisable to keep older versions).

The result (obtained after long series of surprising simplifications)

$$p_2 = \frac{4 \text{Li}_2\left(\frac{7}{2}-\frac{3 \sqrt{5}}{2}\right)}{\sqrt{5}}-\frac{4 \text{Li}_2\left(\frac{1}{2} \left(-3+\sqrt{5}\right)\right)}{\sqrt{5}}+\frac{2 \log \left(\frac{1}{2} \left(\sqrt{5}-1\right)\right) \log \left(25 \left(9-4 \sqrt{5}\right)\right)}{\sqrt{5}}\tag{3c}$$

Numerically,

$$N(p_2) \simeq 0.7556490761416742$$

In Terms of the golden ratio we have

$$p_2 = \frac{2}{\sqrt{5}}(\log (\phi -1) \log (25 (13-8 \phi ))-2 \text{Li}_2(\phi -2)+2 \text{Li}_2(5-3 \phi ))\tag{3d}$$

Technical remarks

Doing the integrals in Mathematica was by no means trivial. I proceeded in the common manner to calculate first the antiderivative and then - after having numerically confirmed continuity - take the limits at $$x=0$$ and $$x=1$$, resp.

While Version 10.1 refused to calculate the Limits at $$x=1$$, Version 8.0 provided results but they contained spurious imaginary parts and terms with products of concealed $$0 * \infty$$ so that the command "Simplify" returned "Indeterminate". A heuristic mixture of numerical checks and tetative cancelleations finally led me to the expressions given which were finally justified numerically.

• Based on the numerical evidence it seems correct with the final two expressions being far more complicated than I could have ever imagined. I am guessing it would also be an almighty herculean task to finally show $p_2 - p_1 = \frac{\pi^2}{5 \sqrt{5}}$. Nice job. – omegadot Feb 11 at 5:03
• @ omagadot You are right, it seems to be hard to prove the simple result for the difference starting from my expressions. Dropping the $(-1)^{n}$ in the sums, and calling them $q$ I find $q_2 - q_1 = \frac{4}{3} \sqrt{3} \Im\left(\text{Li}_2\left(\frac{1}{4} \left(1+i \sqrt{3}\right)\right)\right)+\frac{1}{9} \left(\sqrt{3} \pi \right) \log (4)+\frac{1}{9} \left(\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{1}{3}\right)\right) \simeq 1.17195$ – Dr. Wolfgang Hintze Feb 11 at 11:27
• @ omegadot In trying to find simplifications I discovered this interesting relation valid for $z\lt 0$: $\text{Li}_2\left(\frac{1}{z}\right)+\text{Li}_2(z)+\frac{1}{2} \log ^2(-z)+\zeta (2)=0$. It is most probably known. – Dr. Wolfgang Hintze Feb 11 at 19:09
• @ omegadot I have continued the simplification in an edit. I thank you for accepting my answer but I'd like to ask you to reopen the question for further discussion. – Dr. Wolfgang Hintze Feb 12 at 5:23
• Thanks to the upvoters, but interestingly the problem gets a much higher reputation (7 Points) than the solution (2 points). So IMHO "reputation" seems to be a different measure than "appreciation". But English is not my native language ... – Dr. Wolfgang Hintze Feb 12 at 5:44