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Someone asked me for a formula for the sum of the harmonic progression. So I did some calculations and gave him an approximate formula:

$$\int_1^n\frac{dx}{x} = \frac{y_1 + y_2}{2} + \frac{y_2 + y_3}{2} + \cdots +\frac{y_{n-1} + y_n}{2}$$ where $y_i$ is $i$th term of the HP $$\ln(n) = \frac{y_1}{2} + y_2 +y_3 + \cdots +\frac{y_n}{2}$$

so

$$\sum_{i=1}^n y_i = \ln(n) + \frac{y_1 + y_n}{2}$$

e.g. $1+1/2+\cdots+1/10 = 2.8525 $

actual result $= 2.9289$

My question is, how to correct this formula?

3 Answers 3

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The partial sums of the harmonic series are called "harmonic numbers." The difference between the nth harmonic number and ln(n) tends to a limit as n increases, and that limit is called Euler's constant or gamma.

There's a great book about all this called Gamma: Exploring Euler's Constant.

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A good quick approximation: $\log(n + 0.5) + \gamma$.

A yet more accurate approximation: $\log(n)+\gamma+1/2n + 1/12n^2-1/120n^4+1/252n^6-1/240n^8+1/132n^{10}-691/32760n^{12}+\cdots$, taking as many terms as is convenient.


Error analysis:

The first formula above has maximum error of $1/24n^2$. uday's first answer has error of about 0.0024 ln n. Mine is a better approximation for all $n\neq6$.

The first formula is also better than uday's revised formula for all $n\ge46$. The expected error in uday's answer is about 0.000016.

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Hey guys.... There is no correct simple general formula for sum to n terms of the series 1+1/2+1/3+1/4+ ............. + 1/n

but the following formula will be a good approximation for sum to n-terms of the above series when n>5

S = log(n+0.5) + 0.5772 + 0.04021/(n*n + 0.8848)

Deviation from the actual value fluctuates but remains relatively low..

so i guess this may be a good approximation

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    it would be interesting to see how you found these numbers if they are not just experimentally determined.2010-10-13
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    @muad: To get close to the best approximation with constants in those places, 1.841 should be about $\exp(\gamma)$, 0.9973 should be about 1, and 0.6184 should be about 1/2.2010-10-13
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    @muad: Hi... i have taken values of n from 1 to 1000000 and found the values of S. Using curvefitting tool in matlab i came to that formula...2010-10-20
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    @charls: Hi... i totally agree with you but for smaller values of n,log(n+0.5)+γ has significant amount of error.. so for about n=30 my formula will give a better approximation...2010-10-20
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    @uday kiran: Your new formula is worse than my simple formula for all n > 45.2010-10-24
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    may be !!!! but my formula is very good even for small values of n as compared to your first formula.. i totally agree with you that for very large values of n your formula is better but not for smaller values(as u said < 45)2010-10-25
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    i should say.. my expression came from your expression only... but to reduce the errors for smaller values of n i have added the third expression, for larger values of n the 3rd term in my expression has very little significance hence it transforms into your first expression....2010-10-25
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    in my formula upto 45 you may use 0.5772 as the middle term and after 45 replace 2nd and 3rd terms with γ.I guess this process will give you the optimum results for all n.2010-10-25