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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

A constructive divergent series

A constructive divergent series

Let $\displaystyle \{x_{n}\}$ be a positive sequence satisfying $\lim_{\ n \rightarrow \infty} x_{n}= \infty$ Prove that the series $\displaystyle \sum_{n=1}^{\infty} (1-\frac{x_{n}}{x_{n+1}})$ is divergent.

Till now no answers have been given. The solution I propose involves an 'infinite product'. It is well known the theorem that establish that the 'infinite product' $\displaystyle \prod_{n=1}^{\infty} (1-a_{n})$ diverges if the 'infinite sum' $\displaystyle \sum_{n=1}^{\infty} a_{n}$ diverges. Less known but equally valid is the converse theorem that establishes that the 'infinite sum' $\displaystyle \sum_{n=1}^{\infty} a_{n}$ where $\displaystyle 0<a_{n}<1$ diverges if the 'infinite product' $\displaystyle \prod_{n=1}^{\infty} (1-a_{n})$ diverges to 0 as You can see here...

http://en.wikipedia.org/wiki/Convergence_tests

Applying the converse theorem and setting $\displaystyle a_{n}= 1- \frac{x_{n}}{x_{n+1}}$ we obtain that...

$\displaystyle \prod_{n=1}^{\infty} (1-a_{n}) = \prod_{n=1}^{\infty} \frac{x_{n}}{x_{n+1}}= \frac{x_{1}}{x_{2}}\ \frac{x_{2}}{x_{3}}\ ...$ (1)

Now the product (1) is 'telescopic' and is...

$\displaystyle p_{k}= \prod_{n=1}^{k} \frac{x_{n}}{x_{n+1}}= \frac{x_{1}}{x_{k+1}}$ (2)

... and, because is $\displaystyle \lim_{\ k \rightarrow \infty} x_{k+1}=\infty$ is also $\displaystyle \lim_{\ k \rightarrow \infty} p_{k}=0$, the product (1) diverges and for the converse theorem the series $\displaystyle \sum_{n=1}^{\infty} (1-\frac{x_{n}}{x_{n+1}})$ diverges...

Kind regards

$\chi$ $\sigma$