How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$?

Question

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.

Answer 1

That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$\frac{3^2}{1+2+3} = \frac{9}{6} = 1.5 \neq \frac{1}{1} + \frac{1}{2} + \frac{1}{3}$$.

The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.

Answer 2

The exact expression for $\displaystyle H_n:=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is not known, but you can estimate $H_n$ as below

Let us consider the area under the curve $\displaystyle \frac{1}{x}$ when $x$ varies from $1$ to $n$.

Now note that $\displaystyle H_{n}-\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n-1}$ is an overestimation of this area by rectangles. See below

And $\displaystyle H_n-1=\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is an underestimation of the area. See below

_{(source: uark.edu)}

Hence $$\large H_n-1<\int_{1}^n\frac{1}{x}dx<H_n-\frac{1}{n}\\ \Rightarrow \ln n+\frac{1}{n}<H_n<\ln n+1$$

Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$\large \lim_{n\rightarrow \infty}\left(H_n-\ln n\right)=\gamma\approx 0.57721566490153286060651209008240243104215933593992…$$ $\gamma$ is called the Euler-Mascheroni constant.