Infi-1 14

Cauchy's condensation test #theorem

\begin{matrix} Let a_{n} \geq 0, a_{n} is monotonically non-increasing \\ \sum a_{n} \to M_{1} ⟺ \sum_{n = 0}^{\infty} 2^{n} a_{2^{n}} \to M_{2} \end{matrix}

\begin{matrix} \sum \frac{1}{n^{p}} \to M_{1} ⟺ p > 1 \\ \sum \frac{1}{n^{p}} \to M_{1} ⟺ \sum 2^{n} \frac{1}{(2^{n})^{p}} \to M_{2} \\ \sum 2^{n} \frac{1}{(2^{n})^{p}} = \sum 2^{n - n p} = \sum (2^{1 - p})^{n} \\ ⟹ \sum \frac{1}{n^{p}} \to M_{1} ⟺ 2^{1 - p} < 1 ⟺ 1 - p < 0 ⟺ p > 1 \end{matrix}

\begin{matrix} \sum | a_{n} | \to M_{1} ⟹ \sum a_{n} \to M_{2} \end{matrix}

\begin{matrix} Let a_{n} be monotonically non-increasing, lim_{n \to \infty} a_{n} = 0 \\ Let \exists M : \forall N \in N : | \sum_{n = 1}^{N} b_{n} | \leq M \\ Then \sum a_{n} b_{n} \to M_{1} \end{matrix}