Applications of the Cauchy Condensation Criterion

We provide some example applications of the Cauchy condensation criterion.# Example: Applications of the Cauchy Condensation Criterion

The generalized harmonic series $\sum \frac{1}{n^\alpha}$ is divergent for $\alpha\le 1$ and convergent for $\alpha > 1.$

By the Cauchy condensation criterion, the series $$\sum_{n=0}^\infty \frac{1}{n^\alpha}$$ is convergent if and only if the series $$\sum_{n=0}^\infty\frac{2^n}{(2^n)^\alpha}$$ is convergent.
Since $\frac{2^n}{(2^n)^\alpha}=2^n\cdot 2^{-n\alpha}=2^{n(1-\alpha)}=\left(\frac{1}{2^{1-\alpha}}\right)^n$ we have two cases:
- Case 1: if $\alpha\le 1$ then $\alpha - 1\le 0$ and $2^{\alpha-1}\le 1$. Therefore $\frac{1}{2^{1-\alpha}}\ge 1.$ and the above harmonic series diverges.
- Case 2: if $\alpha > 1$ then $\alpha - 1 > 0$ and $2^{\alpha-1} > 1$. Therefore $\frac{1}{2^{1-\alpha}} < 1$ and the above harmonic series converges, since it is the convergent geometric series.

The series $\sum \frac{1}{n(\log(n))^\alpha}$ is divergent for $\alpha\le 1$ and convergent for $\alpha > 1.$

By the Cauchy condensation criterion, the series $$\sum_{n=0}^\infty \frac{1}{n(\log(n))^\alpha}$$ is convergent if and only if the series $$\sum_{n=0}^\infty\frac{2^n}{2^n(\log(2^n))^\alpha}$$ is convergent.
Since $\frac{2^n}{2^n(\log(2^n))^\alpha}=\frac{1}{(n\log(2))^\alpha}=\log(2)^{-\alpha}\cdot \frac {1}{n^\alpha},$ the convergence behavior is the same as in the case 1 except the constant value $\log(2)^{-\alpha}.$

Thank you to the contributors under CC BY-SA 4.0!

Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition