# 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?

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

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.

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.