Sum of Reciprocals is Divergent
Contents |
Theorem
The harmonic series:
- $\displaystyle \sum_{n=1}^\infty \frac 1 n$
Proof 1
- $\displaystyle \sum_{n=1}^\infty \frac 1 n = \underbrace{1}_{s_0} + \underbrace{\frac 1 2 + \frac 1 3}_{s_1} + \underbrace{\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$
where $\displaystyle s_k = \sum_{i=2^k}^{2^{k+1}-1} \frac 1 i$
From Ordering of Reciprocals, $\forall m < n: \dfrac 1 m > \dfrac 1 n$, so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k+1}}$.
The number of summands in a given $s_k$ is $2^{k+1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:
- $s_k > \dfrac{2^k}{2^{1+k}} = \dfrac 1 2$
Hence the harmonic sum:
- $\displaystyle \sum_{n=1}^\infty \frac 1 n = \sum_{k=0}^\infty \left({s_k}\right) > \sum_{a=1}^\infty \frac 1 2$
the last of which diverges, from the Nth Term Test.
The result follows from the the Comparison Test for Divergence.
$\blacksquare$
Proof 2
Observe that all the terms of the harmonic series are strictly positive.
To prove that the sequence is decreasing, consider:
$f: \R_{>0} \to \R$: $x \mapsto x^{-1}$
From the Power Rule for Derivatives:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm d}{\mathrm dx} x^{-1}\) | \(=\) | \(\displaystyle -x^{-2}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Because $-x^{-2} < 0$ for all $x$ considered, from Derivative of Monotone Function, $f$ is decreasing.
As $f\left({n}\right)$ agrees with the harmonic series for all $n$ in the domain of $\sum \frac 1 n$, we conclude from Monotonicity of Real Sequences that the series is decreasing.
Hence the Cauchy Condensation Test can be applied, and we examine the convergence of:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{n=1}^{\infty}2^n \frac 1 {2^n}\) | \(=\) | \(\displaystyle \sum_{n=1}^{\infty} 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
This diverges, from the Nth Term Test.
Hence $\displaystyle \sum \frac 1 n$ also diverges.
$\blacksquare$
Proof 3
We have that the Integral of Reciprocal is Divergent.
From the Integral Test, the harmonic series also diverges.
$\blacksquare$
