TITLE: Another way of expressing $\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$
QUESTION [13 upvotes]: 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?

REPLY [8 votes]: 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