# Proof that Harmonic numbers Hn = 1+1/2+1/3+...+1/n can not be calculated in constant number of arithmetic operations ?

Consider the basic arithmetical operations of addition, subtraction, multiplication and division.

Can the sum H_n = 1+1/2+1/3+..+1/n be expressed in number of these basic operations that does not depend on n ? Supposedly this can not be done but what is the proof ?

For example the sum:

S_n = 1+2+3+...+n with n-1 additions can be expressed compactly by Gauss formula as n*(n+1)/2 where the number of operations is three and (so it does not depend on n).

I need a similar thing for H_n, or a proof that this can not be done.

Can the sum H_n = 1+1/2+1/3+..+1/n be expressed in number of these basic operations that does not depend on n ? Supposedly this can not be done but what is the proof ?

For example the sum:

S_n = 1+2+3+...+n with n-1 additions can be expressed compactly by Gauss formula as n*(n+1)/2 where the number of operations is three and (so it does not depend on n).

I need a similar thing for H_n, or a proof that this can not be done.

## All Answers (15)

Arturo Ortiz Tapia· Instituto Mexicano del PetroleoPeter Scheiblechner· Lucerne University of Applied Sciences and ArtsArturo Ortiz Tapia· Instituto Mexicano del PetroleoMichael Dreher· Heriot-Watt UniversityDeleted· Ain Shams UniversityPraveen Venkataramana· Massachusetts Institute of TechnologyNita H. Shah· Gujarat UniversityKonrad Burnik· University of ZagrebMy impression is that most of you didn't read the question, but answered by what first came to mind.

The closest thing to a proof that this can not be done in constant operations would be the failure of Gosper's algorithm on input H_n. What I ask is a more elementary type of proof.

Michael Dreher· Heriot-Watt UniversityHowever, such a rational function has the wrong growth behaviour for large n, because H_n growths logarithmically in n.

So, the only chance that a finite number (independent of n) elementary operations can evaluate H_n would imply that you also allow "if-then-else" constructs (or similar things which can modify the flow of the instructions). Analytically, such if-then-else constructs could by written using the Heaviside step function, and if only a finite number of such step functions are allowed, then (here I guess) for large n the algorithm will still be a rational function of n.

Konrad Burnik· University of ZagrebI think this is the answer I am looking for, even though I would like to see this part in full detail:

"...a rational function has the wrong growth behaviour for large n, because H_n growths logarithmically in n"

Praveen Venkataramana· Massachusetts Institute of TechnologyMichael Dreher· Heriot-Watt UniversityYuri Dimitrov· Rousse UniversityAhmad SabihiCan you help by adding an answer?