WolframAlpha says that $\sum^n_{k-1}\left\lfloor \frac{n}{k} \right\rfloor \phi(k) \not = \sum^n_{k=1}k$ which is not true

Question

I submitted the following query to WolframAlpha:

Is sum Floor[n/k]Phi[k], k=1 to n equal to sum k, k to n

and the result (link) was:

$\sum^n_{k-1}\left\lfloor \frac{n}{k} \right\rfloor \phi(k)$ is not always equal to $\sum^n_{k=1}k$

The interesting thing is that it is not true according to the solutions presented in:

• Euler phi Function and Floor Function.
• Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$.

What happened? Is it a bug, or there are some edge cases that WolframAlpha is taking under account?

Answer 1

I've submitted this bug to the WolframAlpha team.

This is their response:

We appreciate your feedback regarding Wolfram|Alpha. The issue you reported has been passed along to our development team for review. Thank you for helping us improve Wolfram|Alpha.