# 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

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:

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

• Mathematica (and therefore WolframAlpha I presume) is not able to simplify the sum down to $n(n+1)/2$ so thats probably the reason it returns false. Returning "I don't know" would probably have been a better answer. – Winther Sep 2 '16 at 15:44
• @Winther This is why I am asking. WolframAlpha should not say that they are not always equal. – Mateusz Piotrowski Sep 2 '16 at 15:45
• It is known that Worlfram alpha makes sometimes mistakes. – N. S. Sep 2 '16 at 15:46
• I think that this mistake was pointed to them couple years back, still not fixed :wolframalpha.com/input/?i=derivative+floor+x – N. S. Sep 2 '16 at 15:48
• @N.S. Interesting. Even the newest Mathematica (version 11) gives strange values for the derivative of the floor function at non-integer values. For example $\text{Floor}'(1.1) = -4.29$ apposed to $0$. – Winther Sep 2 '16 at 16:08

## 1 Answer

I've submitted this bug to the WolframAlpha team.

This is their response:

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