First Derivative of a Summation $\frac{k}{n}\sum_{k}^{n-1}\frac{1}{i}$
What is the first derivative of this with respect to k?
Thank you
 A: There is an alternate definition of the Harmonic Numbers
$$
H_n=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+n}\right)\tag{1}
$$
This agrees with the standard definition when $n\in\mathbb{Z}$, and is analytic except at the negative integers. Furthermore, we get
$$
H_n'=\sum_{k=1}^\infty\frac1{(k+n)^2}\tag{2}
$$
If we notice that
$$
\sum_{j=k}^{n-1}\frac1j=H_{n-1}-H_{k-1}\tag{3}
$$
we get the derivative with respect to $k$ to be
$$
\begin{align}
\frac{\partial}{\partial k}\sum_{j=k}^{n-1}\frac1j
&=-H_{k-1}'\\
&=-\sum_{j=0}^\infty\frac1{(j+k)^2}\tag{4}
\end{align}
$$
Using $(4)$ and the product rule should give the derivative in the question.
A: If you interpret the sum for non-integer (k) like:
$\begin{align}
  \sum_{k \le i \le n - 1} = H_{n - 1} - H_{\lfloor k \rfloor}
\end{align}$
the derivative is just:
$\begin{align}
\begin{cases}
  \frac{1}{n} \sum_{k \le i \le n - 1}
    = \frac{1}{n} (H_{n - 1} - H_{\lfloor k \rfloor})
       & \text{if \(k\) is not an integer} \\
  \text{undefined} & \text{otherwise}
\end{cases}
\end{align}$
But that is certainly a not-very-standard interpretation of your sum notation...
