Question

How do you find the sum of the harmonic series?

Answer 1

Hint: This problem deals with finding the sum of the harmonic series, first of all a harmonic progression is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain 0. The formula to calculate the harmonic mean is given by:
$ \Rightarrow n = \dfrac{1}{{\left[ {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} + .....} \right]}}$

Complete step-by-step answer:
To determine whether the sum of series will converge or diverge, we must use the alternating series test. The test states that for a given series where or where for all $n$, if and is a decreasing sequence, then is convergent.
For a convergent series, the limit of the sequence of partial sums is a finite number. We say the series diverges if the limit is plus or minus infinity, or if the limit does not exist.
Consider the sum of the harmonic series:
$ \Rightarrow \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} + \cdot \cdot \cdot $
By comparison test, and grouping the terms:
$ \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) + \cdot \cdot \cdot $
Now replacing the terms in each group by the smallest term in the group:
$ \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) + \cdot \cdot \cdot $
$ \Rightarrow \sum\limits_{n = 1}^\infty {\dfrac{1}{n}} > 1 + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \cdot \cdot \cdot $
Here there are infinitely many $\dfrac{1}{2}$’s, hence :
$ \Rightarrow \sum\limits_{n = 1}^\infty {\dfrac{1}{n}} = \infty $
The harmonic series diverges because the sequence of partial sums goes to infinity.
The harmonic series is larger than the divergent series, we conclude that harmonic series is also divergent by the comparison test.

Final Answer: $\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} = \infty $

Note:
Please note that a harmonic progression (or a harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression. Equivalently, a sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms.