Question

How do you find the sum of the harmonic series?

Answer 1

We have to find the sum of the harmonic series. The general formula of the terms of the harmonic series is given by $\dfrac{1}{n}$. So, the sum is represented by $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}$. We will write the terms of the harmonic series and regroup it and write the terms as the sum of the lowest term of that group. Continuing this, we will see that the series now has the same ratio added till infinity and so solving which we get out value.

Complete step by step solution:
According to the given question, we are asked to find the sum of the harmonic series.
Harmonic series can be said to be a series of frequencies or overtones which is a multiple of a fundamental tone.
So, from the name itself we can understand that it is a musical term.
The series can be written as the sum of frequencies starting from 1 to infinity but in inverse, that is, the general form of the term is $\dfrac{1}{n}$.
Sum of the harmonic series can be written as: $1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...\infty$
And we have to keep this in mind that the harmonic series diverges. When we say that a series diverges, it means that the series does not approach a value or in other words does not converge and it goes off towards the infinities (either positive or negative), and without coming down on any particular value.
Let us take the harmonic series as,
$\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}=}1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}...\infty$
We will now group the terms in such a way that in each group there is one power of 2. So, it will easier to compute. We have,
$\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n}=}1+\dfrac{1}{2}+\left( \dfrac{1}{3}+\dfrac{1}{4} \right)+\left( \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8} \right)...\infty$
We can now assign the values in the groups as the sum of the lowest term in that group, we have,
$\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}>1+\dfrac{1}{2}+\left( \dfrac{1}{4}+\dfrac{1}{4} \right)+\left( \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \right)...\infty$
If we continue the above process, we will get the sum of the terms as,
$\Rightarrow 1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+...\infty$
We can equalize the above sum to infinity, since the similar ratio is reiterated till infinity, that is,
$\Rightarrow 1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+...\infty =\infty$
Since, the value is infinity, we can confirm it that the harmonic series is divergent in nature.
Therefore, the harmonic series is divergent in nature and thus approaches infinity.

Note: The harmonic series cannot have a fixed value since it is divergent in nature and so will always vary between positive and negative infinity. Also, the above solution can also be used as a proof to show that the harmonic series is divergent in nature.