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$\frac{k}{n}\sum_{k}^{n-1}\frac{1}{i}$

What is the first derivative of this with respect to k?

Thank you

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  • 3
    $\begingroup$ If $k$ is an integer ( as it seams) how can you define a derivative? $\endgroup$ – Emilio Novati Oct 20 '15 at 20:41
  • $\begingroup$ is your sum of the form: (1/k)(1/k+1/(k+1)+1/(k+2)+...+1/(k+n-1))? It is important to be clear! $\endgroup$ – NoChance Oct 20 '15 at 21:38
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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.

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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...

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