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The proof is from the book Advanced Calculus: An Introduction to Linear Analysis by Leonard F. Richardson.

(Ratio Test): Suppose $x_k>0$ for all $k$ and suppose $$\frac{x_{k+1}}{x_k} \rightarrow L$$

as $k \rightarrow \infty$. Then we have the following conclusion.

i. If $L>1$, then $\sum_{k=1}^{\infty}x_k$ diverges.

Proof: If $L>1$, then there exists $K\in\mathbb{N}$ such that $k\geq K$ implies $\frac{x_{k+1}}{x_k}>1$, which implies $0

I can't understand how the inequality $0

$$\frac{x_{k+1}}{x_k}>1$$ $$x_{k+1}>x_k>0$$

and since $x_k$ is an increasing sequence, $x_k \not\rightarrow 0$ and $n$th term test follows.

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    see here i hope this will help http://blogs.ubc.ca/infiniteseriesmodule/appendices/proof-of-the-ratio-test/proof-of-the-ratio-test/ – 2017-01-28
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    It follows from the epsilon-delta definition of a limit. Basically, we end up with $$\left|\frac{x_{k+1}}{x_k}-L\right|<\epsilon$$ – 2017-01-28

1 Answers 1

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Either you messed up the letters of that proof has, imo, a typo. We can say that

$$\lim_{k\to\infty}\frac{x_{k+1}}{x_k}=L>1\implies \exists\,K\in\Bbb N\;\;s.t.\;\;k>K\implies \frac{x_{k+1}}{x_k}>\ell$$

where if $\;L-1=\epsilon>0\;$ , then we can take $\;\ell=L-\frac\epsilon2>1\;$, and then

$$x_{k+1}>x_k\ell\implies \lim_{k\to\infty}x_k\neq0\;\;\text{since}\;\;\{x_k\}\;$$

si a positive sequence.