# Harmonic number of $r$ with order $r$ [closed]

Mathematica recently offered me this expression as the result of an evaluation: HarmonicNumber[r, -r]. I initially thought this must be equivalent to Sum[1/k^r, {k, 1, r}], but it appears I'm wrong:

TrueQ[HarmonicNumber[r, -r] == Sum[1/k^r, {k, 1, r}]]


False

Can someone please tell me what the equivalent of HarmonicNumber[r, -r] is when written in Sum form? I'm sure it's a dumb question (and I'm sure I could ask MMA to do it for me, if I only knew how), but I'd appreciate an answer nonetheless.

## closed as off-topic by Szabolcs, Bob Hanlon, m_goldberg, eyorble, chrisDec 12 '18 at 5:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Bob Hanlon, m_goldberg, eyorble, chris
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• That's not what TrueQ is for. Please check its documentation. – Szabolcs Dec 11 '18 at 16:24
• Have you looked under Details and Options in the documentation of HarmonicNumber? The sum form is given there for any integer input value. – Szabolcs Dec 11 '18 at 16:28
• OK I will, and thank you. But can you please tell me how I can find what HarmonicNumber[r, -r] means? – Richard Burke-Ward Dec 11 '18 at 16:29

HarmonicNumber[r, -r] == Sum[1/k^-r, {k, 1, r}]