Closed form of $\sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right), \quad x \in (0,1)$ Is there a known closed form of the series below?
$$\sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right), \quad x \in (0,1]$$
 A: Denote the sum by $S(x)$. It is straightforward to check by direct differentiation that
$$S'(x)-\frac{S(x)}{x}=x\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2+nx}=
\frac{\psi\left(\frac{x+1}{2}\right) -\psi\left(\frac{x+2}{2}\right)}{2}+\ln 2.$$
The solution of this differential equation satisfying appropriate initial conditions at $x=0$ is
$$S(x)=\frac x2 \int_0^x\left[\psi\left(\frac{s+1}{2}\right) -\psi\left(\frac{s+2}{2}\right)+2\ln 2\right]\frac{ds}{s}.$$
However it does not seem to me that this antiderivative has any chances to be computed in closed form.
A: First, we notice that:
$$\sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right)=x \sum_{n=1}^\infty(-1)^{n+1} \frac 1n \ln\left(1+\frac xn\right)$$
So we only need to find:
$$f(x)=\sum_{n=1}^\infty(-1)^{n+1} \frac 1n \ln\left(1+\frac xn\right)$$
There are several ways to approach this.
1) To get a very good sequence of approximations, we remember the limit:
$$\lim_{n \to \infty} \left(1+\frac xn\right)^n=e^x$$
So, if we truncate the sum at some $N$ big enough, the approximation will be:
$$f_N(x) = \sum_{n=1}^N(-1)^{n+1} \frac 1n \ln\left(1+\frac xn\right)+ \sum_{n=N+1}^\infty(-1)^{n+1} \frac{x}{n^2}$$

$$f_N(x) =\sum_{n=1}^N(-1)^{n+1} \frac 1n \left( \ln\left(1+\frac xn\right)-\frac xn \right)+\frac{\pi^2}{12}x$$

For example:
$$f_2(x)=\ln (1+x)-x-\frac{1}{2} \ln\left(1+\frac{x}{2} \right)+\frac{x}{4}+\frac{\pi^2}{12}x$$
See the plot below, for $f_2(x)$ and $f_3(x)$. The approximations are very good.

Moreover, they provide the bounds for the function:
$$f_3(x)<f(x)<f_2(x)$$
$$f_{2N+1}(x)<f(x)<f_{2N}(x)$$

2) We can also try to use $\ln$ series expansion to transform our sum and get Taylor series for $f(x)$ for $|x|<1$:

$$f(x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \left(1-\frac{1}{2^k} \right) \zeta(k+1) x^k$$

Here $\zeta$ is the Riemann zeta function. This series has slower convergence than the original series, but it is better suited to define the function.
I leave the derivation out, note that I used the fact:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{k+1}}=\left(1-\frac{1}{2^k} \right) \zeta(k+1)$$

It may be interesting to notice, that this sum (like any sum of logarithms) can be connected to an infinite product:
$$e^{f(x)}=\dfrac{1+x}{\sqrt{1+\dfrac{x}{2}}} \cdot \dfrac{\sqrt[3]{1+\dfrac{x}{3}}}{\sqrt[4]{1+\dfrac{x}{4}}} \cdot \dfrac{\sqrt[5]{1+\dfrac{x}{5}}}{\sqrt[6]{1+\dfrac{x}{6}}} \cdots=\prod_{n=1}^{\infty} \dfrac{\sqrt[2n-1]{1+\dfrac{x}{2n-1}}}{\sqrt[2n]{1+\dfrac{x}{2n}}}$$
A: After some calculations I've come across an approximation of a closed form formula: $$f(x) = \sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right), \quad x \in (0,1]\\f(x) \approx \frac{x^{\sqrt3}}{\sqrt3+\frac1a}-\frac1b, \quad a=16, b= 396$$
which yealds to an average error of $1,5084987787$, minimum error $0,0129815729
$of and a maximum of $6,9016346844$.
For further details take a look at the table below:

