Math Challenges

Submissions for Problem #59

Problem #59

\[ \displaylines{\text{Suppose }a_{n}\gt0,s_{n}=a_1+\cdots+a_{n},\text{ and }\sum a_{n}\text{ diverges.}\\ 1)\text{Prove that }\sum\frac{a_{n}}{1+a_{n}}\text{ diverges.}\\ 2)\text{Prove that }\frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}}\ge1-\frac{s_{N}}{s_{N+k}}\\ \text{and deduce that }\sum\frac{a_{n}}{s_{n}}\text{ diverges.}\\ 3)\text{Prove that }\frac{a_{n}}{s_{n}^2}\le\frac{1}{s_{n-1}}-\frac{1}{s_{n}}\\ \text{and deduce that }\sum\frac{a_{n}}{s_{n}^2}\text{ converges.}} \]

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zapwai

Solution: \( \displaylines{1)\text{If }a_{n}\ge1\text{ infinitely often, then }\frac{1}{a_{n}}\le1\implies\frac{1}{a_{n}}+1\le2\text{ and }\frac{a_{n}}{1+a_{n}}\ge\frac12\\ \text{Thus }\sum\frac{a_{n}}{1+a_{n}}\text{ diverges. If instead }a_{n}\lt1\text{ for large n, then}\\ \sum\frac{a_{n}}{1+a_{n}}\gt\frac12\sum_{}^{}a_{n}\text{ which is also divergent.}\\ 2)s_{N}\text{ is increasing monotonically, since }a_{n}\gt0.\text{ Thus }s_1\le s_2\le s_3\le\ldots\\ \text{We therefore have }\frac{1}{s_1}\ge\frac{1}{s_2}\ge\frac{1}{s_3}\ge\ldots\text{ and therefore}\\ \frac{a_{N+1}}{S_{N+1}}+\frac{a_{N+2}}{S_{N+2}}+\cdots+\frac{a_{N+k}}{S_{N+k}}\ge\left(a_{N+1}+a_{N+2}+\cdots a_{N+k}\right)\cdot\frac{1}{S_{N+k}}=\frac{S_{N+k}-S_{N}}{S_{N+k}}\\ \\ \text{which proves the inequality. }k\text{ can be chosen so that }S_{N+k}\ge2S_{N}\text{ as the }S_{N}\text{ are increasing without bound.}\\ \text{Therefore the sequence is not Cauchy, as }\frac{S_{N}}{S_{N+k}}\le\frac12\implies1-\frac{S_{N}}{S_{N+k}}\ge\frac12\\ 3)\text{The right hand side is }\frac{S_{n}-S_{n-1}}{S_{n}\cdot S_{n-1}}=\frac{a_{n}}{S_{n}\cdot S_{n-1}}\ge\frac{a_{n}}{S_{n}^{}\cdot S_{n}}\\ \text{As the right hand side is telescoping, we get a partial sum of }\frac{1}{a_0}-\frac{1}{S_{n}}\to\frac{1}{a_0}\text{ as }n\to\infty} \)