# Proof

(related to Proposition: Cauchy Condensation Criterion)

By hypothesis, $\sum_{k=0}^\infty x_k$ is an infinite series $\sum_{k=0}^\infty x_k$ with a monotonically decreasing real sequence $(x_k)_{k\in\mathbb N}$ of non-negative members $x_k\ge 0.$

### "$\Rightarrow$"

• Assume, $\sum_{k=0}^\infty x_k$ is convergent.
• By definition, there is a real number $s:=\sum_{k=0}^\infty x_k.$
• Since $x_k\ge 0$ and $x_k \ge x_{k+1}$ for all $k\in\mathbb N,$ we have $$\begin{array}{rcl}s&\ge&x_1+x_2+(x_3+x_4)+(x_5+\cdots+x_8)+\cdots+(x_{2^{k-1}+1}+\cdots+x_{2^k})\\&\ge& \frac{x_1}{2}+x_2+2x_4+4x_8+\cdots+2^{k-1}x_{2^k}.\end{array}$$
• It follows $$s_k:=x_1+2x_2+4x_4+16x_8+\cdots+2x_{2^k}\le 2s$$
• Therefore, the real sequence of partial sums of the "condensed series" $s_k=\sum_{n=0}^k 2^nx_{2^n}$ consists of non-negative terms and constitutes a bounded sequence.
• By the monotony criterion, the "condensed series" $$\sum_{n=0}^\infty 2^n x_{2^n}$$ is convergent.

### "$\Leftarrow$"

• Now, assume $\sum_{n=0}^\infty 2^n x_{2^n}$ is convergent and let $t:=\sum_{n=0}^\infty 2^n x_{2^n}.$
• If $2^k\ge n$, then $$\begin{array}{rcl}x_1+x_2+\ldots+x_n&\le&x_1+(x_2+x_3)+(x_4+\cdots+x_7)+\cdots+(x_{2^k}+\cdots+x_{2^{k+1}-1})\\ &\le & x_1+2x_2+4x_4+\cdots+2^kx_{2^k}\\&\le& t.\end{array}$$
• Therefore, the real sequence of partial sums of the series $u_n=\sum_{k=0}^n x_k$ consists of non-negative terms and constitutes a bounded sequence.
• By the monotony criterion, the series $$\sum_{k=0}^\infty x_k$$ is convergent.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
2. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition