Euler's Proof of the Divergence of the Harmonic Series
Theorem
The harmonic series diverges.
$$ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } }=\infty $$
Description
At first glance, the harmonic series appears as if it would converge since its terms continue to decrease in value. However, Oresme elegantly and simply proved that it diverges. This fact is often used as an example to explain the concept of absolute convergence, where the alternating harmonic series converges as shown in $\displaystyle \sum_{n=1}^{\infty} {{(-1)^{n-1}} \over {n}} = 1- {1 \over 2} + { 1 \over 3} - { 1 \over 4 }+ \cdots = \ln 2 < \infty$, whereas the series of their absolute values, the harmonic series, holds true for $\displaystyle \sum_{n=1}^{\infty} \left| {{(-1)^{n-1}} \over {n}} \right| = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } }=\infty$. Therefore, it serves as the simplest example to demonstrate that convergence does not necessarily imply absolute convergence. It also provides a counterexample to show that the inverse of the proposition “If an infinite series converges, then its sequence converges to 0” does not hold.
Proof
The essence of the proof lies in the fact that the sum of infinitely many $1/2$ diverges, and yet, the harmonic series is greater than this sum. The proof is straightforward and neat.
$$ \begin{align*} & \frac { 1 }{ 1 }+\frac { 1 }{ 2 }+\frac { 1 }{ 3 }+\frac { 1 }{ 4 }+\frac { 1 }{ 5 }+\frac { 1 }{ 6 }+\frac { 1 }{ 7 }+\frac { 1 }{ 8 }+\frac { 1 }{ 9 }+\cdots \\ =& \frac { 1 }{ 1 }+\frac { 1 }{ 2 }+\left( \frac { 1 }{ 3 }+\frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 5 }+\frac { 1 }{ 6 }+\frac { 1 }{ 7 }+\frac { 1 }{ 8 } \right) +\frac { 1 }{ 9 }+\cdots \\ >& 1+\frac { 1 }{ 2 }+\left( \frac { 1 }{ 4 }+\frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 8 }+\frac { 1 }{ 8 }+\frac { 1 }{ 8 }+\frac { 1 }{ 8 } \right) +\cdots \\ =& 1+\frac { 1 }{ 2 }+\frac { 1 }{ 2 }+\frac { 1 }{ 2 }+\cdots \end{align*} $$
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