<s>[INST] How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$
$\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P.
Therefore the sum of the series can be written as :
$\Rightarrow \frac{(3)^3}{(A+B+C)}$
Is this correct? Please suggest.
 [/INST] There are other ways to represent the harmonic series, for example, recall the geometric sum:
$$\frac{1-r^n}{1-r}=1+r+r^2+\dots+r^{n-1}$$
And integrate both sides with respect to $r$ from $0$ to $1$:
$$\begin{align}\int_0^1\frac{1-r^n}{1-r}dr&=\int_0^11+r+r^2+\dots+r^{n-1}dr\\&=\left.\frac11r+\frac12r^2+\frac13r^3+\dots+\frac1nr^n\right|_0^1\\&=\frac11+\frac12+\frac13+\dots+\frac1n\end{align}$$
So we can rewrite the harmonic series as
$$1+\frac12+\frac13+\dots+\frac1n=\int_0^1\frac{1-r^n}{1-r}dr$$
Which is most useful for deriving many formulas.  I used in deriving a representation of the gamma function, for example.
Note that:
$$\ln(\Gamma(n+1))=\ln(n\Gamma(n))=\ln(\Gamma(n))+\ln(n)$$
Take the derivative of both sides to get
$$\psi(n+1)=\psi(n)+\frac1n$$
Repeated application of this formula gives:
$$\begin{align}\psi(n+1)&=\psi(n)+\frac1n\\&=\psi(n-1)+\frac1{n-1}+\frac1n\\&=\psi(1)+1+\frac12+\frac13+\dots+\frac1n\\\psi(n+1)+\gamma&=1+\frac12+\frac13+\dots+\frac1n\end{align}$$
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