In mathematics, harmonic series is an infinite series. That is, the partial sums attained by adding the successive terms grow without limit, or, placed an additional way, this sum will infinity.

In sequence and series, we deal with various types of series and sequences such as geometric series and sequences and arithmetic series and sequences. In this article will learn about harmonic series in detail.

The harmonic series is one of the most important infinite series of mathematics. This is the sum of reciprocals of the positive integers. Harmonic series is represented as below:

$\sum_{n=1}^{\infty}$ $\frac{1}{n}$ = $\frac{1}{1}$ + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ................
Harmonic series is purely divergence series, which diverges to infinity. We can prove this using integral test and comparison test.

Consider nth partial summation notation of harmonic series:

$\sum_{k=1}^n$ $\frac{1}{k}$ = 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + ......+$\frac{1}{n}$

As n tends to infinity, its sum diverges to infinity.

i.e. $\sum_{k=1}^{\infty}$ $\frac{1}{k}$ = 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + ......
= $\infty$

The partial sum formula for the harmonic series is as:

$\sum_{n=1}^{\infty}$ $\frac{1}{n}$ = $\frac{1}{1}$ + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ................ = 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ................

= $\frac{n(n+1)}{2}$
This concept can be proved by various ways, here will discuss about two tests: comparison test and integral test.

Integral Test:
Let us prove the harmonic series diverges by comparing its sum with an improper integral. Consider an arrangement of rectangle, in which each rectangle is $\frac{1}{n}$ units high and 1 unit wide.

Harmonic Series

The total area of the rectangles = Sum of the harmonic series

Area of Rectangle = 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + .....

Area under the curve (marked in blue) = $\int_1^{\infty}$ y dx

= $\int_1^{\infty}$ $\frac{1}{x}$ dx = $\infty$

=> Total area of the rectangles = $\infty$

=>  $\sum_{n=1}^k$ $\frac{1}{n}$ > $\int_1^{k+1}$$\frac{1}{x}$ dx = ln(k + 1)

Comparison Test:

Compare harmonic series with the another divergent series.

1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + $\frac{1}{6}$ + $\frac{1}{7}$ + $\frac{1}{8}$ + ................

> 1 + $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + ................

Just about every term with the harmonic series is in excess of or maybe adequate to the actual equivalent term with the second series, and then the amount of the actual harmonic series have to be in excess of the sum of the 2nd series. Nevertheless, the sum of the 2nd series is infinite:

1 + $\frac{1}{2}$ + $(\frac{1}{4} + \frac{1}{4})$ $(\frac{1}{8} + \frac{1}{8}+\frac{1}{8}+\frac{1}{8})$ + .............

= 1 + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + ..........

= $\infty$
Which shows that series is diverges.

Comparison test shows that $\sum_{n=1}^{2^k}$ $\frac{1}{n}$ $\geq$ 1 + $\frac{k}{2}$ $\forall$ k$^+$.
If we compare series of the form $\int_1^{\infty}$ $\frac{1}{x^k}$ dx to the series

$\sum_{n=1}^{\infty}$ $\frac{1}{n^k}$, we get the conclusion is

- Series converges if p > 1

- Series diverges if p $\leq$ 1

Since harmonic series is the sum of reciprocals of the positive integers and diverges always.

The harmonic series does not converge, the alternating harmonic series does converge due to the alternation of signs.
Below are examples on Harmonic series:

Example 1: Show that series $\sum_{k=4}^{\infty}$ $\frac{1}{k}$ is divergent.


Harmonic series is defined as: $\sum_{k=1}^{\infty}$ $\frac{1}{k}$
or $\sum_{k=1}^{\infty}$ $\frac{1}{k}$ = 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\sum_{k=4}^{\infty}$ $\frac{1}{k}$

(Means, if we add first 3 terms of harmonic series in the given series, will get harmonic series)

=> $\sum_{k=4}^{\infty}$ $\frac{1}{k}$ = $\sum_{k=1}^{\infty}$ $\frac{1}{k}$ - $\frac{11}{6}$

 = $\infty$ - $\frac{11}{6}$

= $\infty$ (If we add or subtract anything in infinity will get result as infinity)

Therefore, the series is divergent.

Example 2: Show that if we multiply a divergent series by a constant it will still be divergent.

Solution: Let us prove this statement by taking an example.

Consider a series as $\sum_{k=1}^{\infty}$ $\frac{20}{k}$

We can also rewrite this as  $\sum_{k=1}^{\infty}$ $\frac{20}{k}$ =

20$\sum_{k=1}^{\infty}$ $\frac{1}{k}$   .....(1)

Series $\sum_{k=1}^{\infty}$ $\frac{1}{k}$ is a harmonic series divergent at infinity.

So $\sum_{k=1}^{\infty}$ $\frac{20}{k}$ = 20 $\times$ $\infty$ = $\infty$

Thus, if we multiply a divergent series by a constant it will still be divergent.