Do you know any nice way of expressing
$$\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$$ ?
Some simple manipulations involving the integrals lead to an expression that also uses
the hypergeometric series. Is there any way of getting a form that doesn't use the HG function?
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal I}_{n} \equiv\sum_{k = 0}^{n}{H_{k + 1} \over n - k + 1} =\sum_{k = 1}^{n + 1}{H_{k} \over n - k + 2}:\ {\large ?}}$.
\begin{align} \sum_{n = 0}^{\infty}{\cal I}_{n}z^{n}& =\sum_{n = 1}^{\infty}{\cal I}_{n - 1}z^{n - 1}= \sum_{n = 1}^{\infty}z^{n - 1}\sum_{k = 1}^{n}{H_{k} \over n - k + 1} =\sum_{k = 1}^{\infty}H_{k}\sum_{n\ =\ k}^{\infty}{z^{n - 1} \over n - k + 1} \\[3mm]&=\sum_{k = 1}^{\infty}H_{k}\sum_{n = 1}^{\infty}{z^{n + k - 2} \over n} ={1 \over z^{2}}\sum_{k = 1}^{\infty}H_{k}z^{k}\sum_{n = 1}^{\infty}{z^{n} \over n} ={1 \over z^{2}}\bracks{-\,{\ln\pars{1 - z} \over 1 - z}}\bracks{-\ln\pars{1 - z}} \\[3mm]&={\ln^{2}\pars{1 - z} \over z^{2}\pars{1 - z}} ={1 \over z^{2}}\, \lim_{\mu\ \to\ -1}\partiald[2]{\pars{1 - z}^{\mu}}{\mu} ={1 \over z^{2}}\, \lim_{\mu\ \to\ -1}\partiald[2]{}{\mu}\sum_{n = 0}^{\infty}\pars{-1}^{n}z^{n} {\mu \choose n} \\[3mm]&={1 \over z^{2}}\, \lim_{\mu\ \to\ -1}\partiald[2]{}{\mu}\sum_{n = 0}^{\infty}z^{n} {-\mu + n - 1\choose n} \end{align}
There's not any contribution from the first two terms such that: \begin{align} \sum_{n = 0}^{\infty}{\cal I}_{n}z^{n}&= {1 \over z^{2}}\,\lim_{\mu\ \to\ -1}\partiald[2]{}{\mu}\sum_{n = 2}^{\infty}z^{n} {-\mu + n - 1\choose n} = \lim_{\mu\ \to\ -1}\partiald[2]{}{\mu}\sum_{n = 0}^{\infty}z^{n} {-\mu + n + 1\choose n + 2} \end{align}
\begin{align} {\cal I}_{n}&\equiv\sum_{k = 0}^{n}{H_{k + 1} \over n - k + 1} =\lim_{\mu\ \to\ -1}\partiald[2]{}{\mu}{-\mu + n + 1\choose n + 2} \\[5mm]&=\lim_{\mu\ \to\ -1}{-\mu + n + 1 \choose n + 2}\left\lbrace% \bracks{\Psi\pars{-\mu + n + 2} - \Psi\pars{-\mu}}^{2} \right. \\&\left.\phantom{\lim_{\mu\ \to\ -1}\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,}- \Psi'\pars{-\mu} + \Psi'\pars{-\mu + n + 2}\right\rbrace \\[5mm]&=\bracks{\Psi\pars{n + 3} - \underbrace{\Psi\pars{1}} _{\ds{=\ \color{#c00000}{-\gamma}}}}^{2} - \underbrace{\Psi'\pars{1}} _{\ds{=\ \color{#c00000}{\pi^{2} \over 6}}} +\Psi'\pars{n + 3} \end{align}
$$\color{#66f}{\large% \sum_{k = 0}^{n}{H_{k + 1} \over n - k + 1} =\bracks{\Psi\pars{n + 3} + \gamma}^{2} + \Psi'\pars{n + 3} - {\pi^{2} \over 6}} $$
$\ds{\Psi\pars{z}}$ and $\ds{\gamma}$ are the Digamma Function and the Euler-Mascheroni Constant, respectively. See this link.
Using small steps:
$$
\begin{align}
\sum_{k=1}^{n+1}\frac{H_k}{n-k+2}
&=\sum_{k=1}^{n+1}\sum_{j=1}^k\frac1j\frac1{n-k+2}\tag{1}\\
&=\sum_{j=1}^{n+1}\sum_{k=j}^{n+1}\frac1j\frac1{n-k+2}\tag{2}\\
&=\sum_{j=1}^{n+1}\sum_{k=j}^{n+1}\frac1j\frac1{k-j+1}\tag{3}\\
&=\sum_{k=1}^{n+1}\sum_{j=1}^k\frac1j\frac1{k-j+1}\tag{4}\\
&=\sum_{k=1}^{n+1}\frac1{k+1}\sum_{j=1}^k\left(\frac1j+\frac1{k-j+1}\right)\tag{5}\\
&=\sum_{k=1}^{n+1}\frac1{k+1}\sum_{j=1}^k\frac2j\tag{6}\\
&=\sum_{k=1}^{n+1}\frac1{k+1}\sum_{j=1}^k\frac1j
+\sum_{j=1}^{n+1}\frac1{j+1}\sum_{k=1}^j\frac1k\tag{7}\\
&=\color{#C00000}{\sum_{k=1}^{n+1}\frac1{k+1}\sum_{j=1}^k\frac1j}
+\color{#00A000}{\sum_{k=1}^{n+1}\frac1k\sum_{j=k}^{n+1}\frac1{j+1}}\\
&+\color{#0000FF}{\sum_{k=1}^{n+2}\frac1{k^2}}-\sum_{k=1}^{n+2}\frac1{k^2}\tag{8}\\
&=\left(\sum_{k=1}^{n+2}\frac1k\right)^2-\sum_{k=1}^{n+2}\frac1{k^2}\tag{9}\\[6pt]
&=H_{n+2}^2-H_{n+2}^{(2)}
\end{align}
$$
Explanation:
$(1)$: $H_k=\sum\limits_{j=1}^k\frac1j$
$(2)$: change order of summation
$(3)$: substitute $k\mapsto n+j+1-k$
$(4)$: change order of summation
$(5)$: $\frac1j\frac1{k-j+1}=\frac1{k+1}\left(\frac1j+\frac1{k-j+1}\right)$
$(6)$: $\sum\limits_{j=1}^k\frac1j=\sum\limits_{j=1}^k\frac1{k-j+1}$
$(7)$: substitute $j\leftrightarrow k$ in the right sum
$(8)$: change order of summation in the green sum; add $0$
$(9)$: red, green, blue sum covers first term bigger, smaller, equal
Yes: $$\begin{eqnarray*}\color{red}{\sum_{k=1}^{n+1}\frac{H_k}{n+2-k}}=\sum_{k=1}^{n+1}\sum_{r+s=k}\frac{1}{rs}=\color{red}{H_{n+2}^2-H_{n+2}^{(2)}.}\end{eqnarray*}$$ For the proof, see this other question. It is line $(4)$ in my second answer.
Let $f(x)=\displaystyle\sum\limits_{n\geqslant 0}\frac{1}{n+1}x^n=-x^{-1}\log(1-x)$
Then $\displaystyle g(x)=\frac{1}{1-x}f(x)=\sum_{n\geqslant 0}\sum_{k=0}^n \frac{1}{k+1}x^n=\sum_{n\geqslant 0}H_{n+1} x^n$ and you want the coefficient of $x^n$ (i.e. $f^{(n)}(0)/n!$) in $$f(x)g(x)=x^{-2}\log^2(1-x)/(1-x)$$ I guess you can do some complex countour integration trickery, which I don't know about.
Note that by the observation, i.e. since it is $[x^n]$ in $(1-x)^{-1}f(x)^2$, we can associate $f(x)^2$ first, and thus $$\sum_{k=0}^n \frac{H_{k+1}}{n-k+1}=\sum_{k=0}^n\sum_{j=0}^k\frac{1}{j+1}\frac{1}{k-j+1}$$
This is a really rough approximation: write $H_{k+1} \sim \log (k+1) \sim \log k + \frac{1}{k}$ then you get two sums: \begin{align} &\sum_{k=1}^{n}\frac{\log k }{n-k+1} + \sum_{k=1}^{n} \frac{1}{k(n-k+1)}\\ &\leq \log n \sum_{k=1}^{n}\frac{1}{n-k+1} + \sum_{k=1}^{n} \frac{1}{k(n-k+1)} \\ &\sim \log^2n + \frac{H_n}{n+1} \end{align}
