Is there a closed form for the harmonic series?

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SUMMARY

The harmonic series does not have a closed form in the traditional sense, as established in the discussion. Euler's approximation, represented as H(n) ~ ln(n) + γ (where γ is the Euler-Mascheroni constant), illustrates asymptotic behavior rather than equality. The summation involving Bernoulli numbers does not qualify as a closed form either. The consensus is that closed forms must adhere to specific definitions of equality and acceptable functions, which the harmonic series does not satisfy.

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  • Understanding of asymptotic notation and equivalence
  • Familiarity with the harmonic series and its properties
  • Knowledge of Euler-Mascheroni constant (γ)
  • Basic concepts of Bernoulli numbers and their applications
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  • Study asymptotic analysis in mathematical series
  • Explore the role of Bernoulli numbers in number theory
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avocadogirl
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Taking from Euler's offering that:

\Sigma1/n= ln(n) + \gamma

could you say that there is a closed form of the harmonic series?

Does Euler's offering qualify?
 
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Your = is not the usual =, but the asymptomic =. As such it is not a closed form in the usual sense. To talk of a closed form one needs to define which functions and operations are to be acceptable and as in this case in what sense two forms are to be equal. The harmonic series in usually not considered to have a closes form as it can not be written in terms of the usual function with the usual =.
 
Could you please explain a little more about this asymptotic equivalence?

Would it be more accurate to include the summation of a "Kth" Bernoulli number, divided by the product of "K" and (n raised to the power of "K"), as K goes from one to infinity?

But, including this summation would not constitute closed form, correct?
 
avocadogirl said:
Taking from Euler's offering that:

\Sigma1/n= ln(n) + \gamma]
No, Euler never said any such thing. The equation you write is obviously false. On the left, you have a sum with n varying over some unstated domain so the sum itself does NOT depend on n, but the right hand side does.

could you say that there is a closed form of the harmonic series?

Does Euler's offering qualify?
 
avocadogirl said:
Could you please explain a little more about this asymptotic equivalence?

Would it be more accurate to include the summation of a "Kth" Bernoulli number, divided by the product of "K" and (n raised to the power of "K"), as K goes from one to infinity?

But, including this summation would not constitute closed form, correct?

H(n)= ln(n) + Euler–Mascheroni
where Hn is the nth harmonic number
would better be written
H(n)~ln(n) + Euler–Mascheroni
to avoid confusion
interpeted as
limit [-H(n)+ ln(n) + Euler–Mascheroni]=0

That is to say that while the two sides are never equal the get closer the large n gets

and no a integral or a sum (finite or infinite) would not be considered a closed form for harmonic numbers
 
Thank you! That clarifies.
 

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