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### The harmonic series

The **harmonic series** is

There is no simple expression for the sum of the first \(n\) terms of this series. Does the series have a limiting sum? The following argument shows that the answer is no.

We can group the terms of the series as follows:

\[ 1 + \dfrac{1}{2} + \Bigl(\dfrac{1}{3} + \dfrac{1}{4}\Bigr) + \Bigl(\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}\Bigr) + \Bigl(\dfrac{1}{9} + \dfrac{1}{10} + \dots + \dfrac{1}{16}\Bigr) + \dotsb. \]Each term is greater than or equal to the last term in its bracket, and so we can write

\begin{align*} 1 + \dfrac{1}{2} &+ \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dotsb \\ &\geq 1 + \dfrac{1}{2} + \Bigl(\dfrac{1}{4} + \dfrac{1}{4}\Bigr) + \Bigl(\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\Bigr) + \Bigl(\dfrac{1}{16} + \dfrac{1}{16} + \dots + \dfrac{1}{16}\Bigr) + \dotsb \\ &= 1 + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dotsb, \end{align*}which grows without bound. So the harmonic series does not have a limiting sum.

On the other hand, if we square each term and look at the series

\[ 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} + \dotsb, \]then it can be shown (although it is not all that easy) that this series has a limiting sum of \(\dfrac{\pi^2}{6}\). This result was proven by Euler in the 18th century.