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Let $(s_n)$ be a sequence of positive terms such that the sequences of ratios $(\frac{s_n+1}{s_n})$ converges to $L$. Prove that if $L>1$, then $\lim s_n=+\infty \!\,$

So I know I have to use the theorem in my book that says "let $(s_n)$ be a sequence of positive numbers. Then $\lim s_n=+\infty \!\,$ iff $\lim(1/s_n)=0$.

So here's my attempt:

Suppose $(s_n)$ is a sequence of positive terms such that the sequences of ratios $\lim(\frac{s_n+1}{s_n})=L$, then (I am not sure if this is even mathematically correct to say, can I just "flip" the thing that we're taking the limit of and then say the limit of this new thing equals $1$Can a house be taken in eminent domain when the government has no intended use for the property? divided by the old limit?) $\lim\left(\frac{s_n}{s_n+1}\right)=1/L$. Since $L>1$, then $1/L<1\dots$ and then I'm stuck..

Michael Hardy
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Math Major
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2 Answers2

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Hint: Let $\varepsilon>0$ such that $L>L-\varepsilon=c>1$.

There exists $N\in \Bbb N$ such that , for all $n\geq N$ ; $L+\varepsilon\geq\dfrac{s_{n+1}}{s_n}\geq L-\varepsilon=c$, so $s_{n+1}\geq cs_n\geq c^{n-N}s_N$ (for $n\geq N$).

Hamou
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Since $s_{n+1}>s_n>0$, for each $n\in \mathbb N$, $(s_n)$ is a monotone increasing sequence and convergence of $s_n \to 0$ is excluded.

Assume know that $(s_n)$ is bounded and that (therefore) there exists $s:=\displaystyle\lim_{n\to \infty}s_n$, with $s\in \mathbb R_+$. Since $\lim s_n=\lim s_{n+1}$ then$$1<L=\lim_{n\to \infty}\frac{s_{n+1}}{s_n}=\frac{\displaystyle\lim_{n\to \infty}s_{n+1}}{\displaystyle\lim_{n\to \infty}s_{n}}=\frac{s}{s}=1$$ which is a contradiction. Thus $(s_n)$ is unbounded and therefore diverges.

Jimmy R.
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