amount of harmonic development?

A person asked me for a formula for the amount of the harmonic development. So I did some estimations and also offered 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 regard to the HP. $$\ln(n) = \frac{y_1}{2} + y_2 +y_3 + \cdots +\frac{y_n}{2}$$

so

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

as an example $1+1/2+\cdots+1/10 = 2.8525 $

real outcome $= 2.9289$

My inquiry is, just how to remedy this formula?

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2019-05-09 11:21:44
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Answers: 3

Hey individuals There is no proper straightforward basic formula for amount to n regards to the collection 1+1/ 2+1/ 3+1/ 4 npls.+1/n

yet the adhering to formula will certainly be an excellent estimate for amount to n - regards to the above collection when n > 5

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

Deviation from the real value rises and fall yet continues to be reasonably reduced.

so i hunch this might be an excellent estimate

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2019-05-21 10:36:42
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An excellent fast estimate: $\log(n + 0.5) + \gamma$.

A yet extra exact estimate: $\log(n)+\gamma+1/2n + 1/12n^2-1/120n^4+1/252n^6-1/240n^8+1/132n^{10}-691/32760n^{12}+\cdots$, taking as several terms as is hassle-free.


Mistake analysis:

The first formula over has maximum mistake of $1/24n^2$. uday is first solution has mistake of concerning 0.0024 ln n. Mine is a far better estimate for all $n\neq6$.

The first formula is additionally far better than uday is changed formula for all $n\ge46$. The anticipated mistake in uday is solution has to do with 0.000016.

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2019-05-21 09:53:58
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The partial amounts of the harmonic collection are called "harmonic numbers." The distinction in between the umpteenth harmonic number and also ln (n) often tends to a restriction as n rises, which restriction is called Euler's constant or gamma.

There's a wonderful publication concerning all this called Gamma: Exploring Euler's Constant.

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2019-05-10 07:10:13
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