Closed form of finite Euler sum $\sum_{k=1}^n \frac{ H_{k}}{(2k+1)}$ Recently (Finite sums of products of harmonic numbers like $\sum_{k=1}^n H_{k} H_{2k}$) I came across this finite Euler sum 
$$p_{2}(n) = \sum_{k=1}^n \frac{H_{k}}{2k+1}\tag{1}$$
and I wonder if it has a closed form.    
Closed form means in this context expressible terms in harmonic numbers, similar to the relation
$$\sum_{k=1}^n H_{k} = (n+1)H_{n}-n\tag{2}$$
The contrary would be that $p_{2}(n)$ is irreducible, or that it belongs to class of irreducible sums.
Writing $H_{k}=\sum_{i=1}^k 1/i$ and interchanging the order of summation we get the relation
$$\sum_{k=1}^n \frac{H_{k}}{2k+1}=\frac{1}{2} H_n H_{n+\frac{1}{2}}-\frac{1}{2}\sum_{k=1}^{n}\frac{H_{k-\frac{1}{2}}}{k}\tag{3}$$
This leads us to define
$$p_{3}(n) = \sum_{k=1}^{n}\frac{H_{k-\frac{1}{2}}}{k}\tag{4}$$
and the question of a closed form has shifted to another sum. 
 A: Lets simplify the sum first:
\begin{align}
\sum_{k=1}^n\frac{H_k}{2k+1}=\sum_{k=0}^n\frac{H_k}{2k+1}=\sum_{k=1}^{n+1}\frac{H_{k-1}}{2k-1}=\color{blue}{\sum_{k=1}^n\frac{H_{k-1}}{2k-1}}+\frac{H_n}{2n+1}\tag{1}
\end{align}
Now apply Abel's summation to the blue sum: 
$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\ $ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $
and by letting let $\ \displaystyle a_k=\frac{1}{2k-1}\ $ , $\ \displaystyle b_k=H_{k-1}$, we get
\begin{align}
\sum_{k=1}^n\frac{H_{k-1}}{2k-1}&=\left(\sum_{i=1}^n\frac1{2i-1}\right)H_n+\sum_{k=1}^n\left(\sum_{i=1}^k\frac1{2i-1}\right)\left(H_{k-1}-H_k\right)\\
&=\left(H_{2n}-\frac12H_n\right)H_n+\sum_{k=1}^n\left(H_{2k}-\frac12H_k\right)\left(-\frac1k\right)\\
&=H_nH_{2n}-\frac12H_n^2-\sum_{k=1}^n\frac{H_{2k}}{k}+\frac12\sum_{k=1}^n\frac{H_k}{k}\\
&=H_nH_{2n}-\frac12H_n^2-\sum_{k=1}^n\frac{H_{2k}}{k}+\frac12\left(\frac12H_n^2+\frac12H_n^{(2)}\right)\\
&=H_nH_{2n}-\frac14H_n^2+\frac14H_n^{(2)}-\sum_{k=1}^n\frac{H_{2k}}{k}\tag{2}
\end{align}
Plugging $(2)$ in $(1)$, we get

$$\sum_{k=1}^n\frac{H_k}{2k+1}=H_nH_{2n}-\frac14H_n^2+\frac14H_n^{(2)}+\frac{H_n}{2n+1}-\sum_{k=1}^n\frac{H_{2k}}{k}$$

Maybe the last sum has a closed form? I hope you find this approach useful.
A: Nothing on the closed form so far, but we can derive a double integral expression for the second sum which might be useful:
$$\sum_{k=1}^n \frac{H_{k-a}}{k}=\int_0^1 \int_0^1 \left(\frac{1-v^n}{1-v}-t^{1-a} \frac{1-(t v)^n}{1-t v} \right) \frac{dv dt}{1-t} \\ a<1$$
In particular:

$$p_3(n)=\int_0^1 \int_0^1 \left(\frac{1-v^n}{1-v}-\sqrt{t} \frac{1-(t v)^n}{1-t v} \right) \frac{dv dt}{1-t}$$

We can also derive a nice looking generating function:
$$G(x)=\sum_{n=1}^\infty p_3(n) x^{n-1}= \frac{1}{1-x} \int_0^1 \int_0^1 \left(\frac{1}{1-xv}- \frac{\sqrt{t}}{1-x t v} \right) \frac{dv dt}{1-t}$$
From which follows:

$$(1-x) G(x)= \sum_{q=0}^\infty \frac{H_{q+1/2}}{q+1} x^q$$

This series appears to have a closed form, as can be seen from checking particular values of $x$ in Wolfram Alpha, such as $x=1/3$ or $x=1/5$.
For example:
$$ \frac23 G \left(\frac13 \right)=\frac32 \left(4 \log^2 2-4 \log 2 \log 3+2 \log 2 \log (\sqrt{3}-1)- \\ -4 \log( \sqrt{3}-1) \log( \sqrt{3}+1)+ \log^2 (\sqrt{3}+1)-\log^2 (\sqrt{3}-1) \right)$$
For $x=1/2$ the closed form contains dilogarithms, but maybe it can be simplified.

If we find a closed form for $G(x)$, then the problem formally reduces to differentiating $G(x)$ and evaluating the derivatives at $x=0$.
Update
With the help of Mathematica I have derived the following simple closed form for the generating function:

$$G(x)= \frac{2}{x(1-x)} \left( \operatorname{arctanh}^2 \sqrt{x}+\log 2 \log (1-x) \right)$$

I have checked a few derivatives and it works.
Note:
$$p_3(n+1)= \frac{1}{n!} \lim_{x \to 0} \frac{\partial^n}{\partial x^n} G(x)$$
Here's an illustration:

A: Generating function of finite sum
I would like to take the opportunity of this self answer to generalize the brilliant idea of Yurij S to calculate the generating function (g.f.) of the sum of interest. I suggest that a closed form of the g.f. of a finite (Euler) sum might be considered as a kind of substitute for the original closed form of the sum which might be lacking.
Here, as an example, we calculate directly the generating function of the original sum 
$$p_2(n)=\sum_{k=1}^n \frac{H_k}{2k+1}$$
hoping to get a closed expression for the g.f. (the hope turns out to be justified).
The g.f. of our sum of interest is defined as
$$g_2(z) = \sum_{k=1}^\infty p_2(n) z^n$$
We proceed in four steps.
Step 1: write sum as double integral
Using the relations $H_k = \int_0^1 \frac{1-x^k}{1-x}\,dx$ and $\frac{1}{2k+1}=\int_0^1 y^{2k}\,dy$ the sum $p_2(n)$ transforms itself naturally into a double integral: 
$$p_2(n) = \int_0^1 \,dx \int_0^1 \,dy \sum_{k=1}^n y^{2k} \frac{1-x^k}{1-x}$$
Step 2: perform the finite sum
Doing the sum under the integral gives the integrand
$$i(n) = \frac{y^2 \left(\left(y^2-1\right) x^{n+1} y^{2 n}-x y^{2 n+2}+y^{2 n}+x-1\right)}{(x-1) \left(y^2-1\right) \left(x y^2-1\right)}$$
Step 3: form the g.f. with the Integrand
Before integrating we first form the g.f. under the integral 
$$i_g(z) = \sum_{n=1}^\infty i(n) z^n = -\frac{y^2 z}{(z-1) \left(y^2 z-1\right) \left(x y^2 z-1\right)}$$
Step 4: do the integration (in an appropriate order)
Now we do the $x$-integral
$$i_{g,x}(z)= \int_0^1 i_g(z) \,dx = -\frac{\log \left(1-y^2 z\right)}{(1-z) \left(1-y^2 z\right)}$$
and finally the $y$-integral gives the g.f.
$$g(z)= \int_0^1 i_{g,x}(z) \,dy = \frac{1}{12 (z-1) \sqrt{z}} \left(-12 \text{Li}_2\left(\frac{\sqrt{z}-1}{\sqrt{z}+1}\right)+12 \tanh ^{-1}\left(\sqrt{z}\right)^2+12 \left(\log (4-4 z)-2 \log \left(\sqrt{z}+1\right)\right) \tanh ^{-1}\left(\sqrt{z}\right)-\pi ^2\right)$$
This is the desired closed form of the g.f.
The series expansion of $g(z)$ about $z=0$ starts like this
$$g(z) = \frac{z}{3} + \frac{19 z^2}{30}+ \frac{94 z^3}{105}+\frac{4259 z^4}{3780}+\frac{2774 z^5}{2079}+...$$
It has only positive integer powers of $z$ as it should, and it is easy to identify the coefficient of $z^n$ as the value of $p_2(n)$ by direct comparison (left as an easy exercise to the reader).
Lemma: relation between generating functions
There is an interesting and useful relation between the g.f. of a series $a(n)$ and the g.f. of a finite sum of the series.
Let
$$g_a(z) = \sum_{n=1}^\infty a(n) z^n$$
$$s(n) = \sum_{k=1}^n a(k)$$
$$g_s(z) = \sum_{n=1}^\infty s(n) z^n$$
then
$$g_s(z) = \frac{1}{1-z} g_a(z)$$
The proof is easily found by interchanging the order of summation in $g_s(z)$
Here are some simple applications of the lemma.
Harmonic number
Taking $a(k)=1/k$ gives the generating function $g_a(z) = \sum_{k=1}^\infty \frac{z^k}{k} = - \log(1-z)$. Hence for the sum we get the g.f. $g_s(z) = \frac{g_a(z)}{1-z} = \frac{ - \log(1-z)}{1-z}$. Noticing thet the sum is just the harmonic number we have recovered the g.f. of the latter.
Iterated sums
Defining the $q$-fold iterated sum recursively by
$s(0,n) = a(n)$, $s(q+1,n) = \sum_{k=1}^n s(q,k)$
the g.f. of $s(q,n)$ is given by the simple formula
$$g(q,z) = \sum_{k=1}^\infty z^n s(q,n) = \frac{g(0,z)}{(1-z)^q}$$
