if $\limsup a_{n+1}/a_n = L < 1$, then $\sum a_n$ converges

Question

I must demonstrate that, $\forall \{a_n\}$ define in $\mathbb{R}^{+}$, if $$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = L < 1 \implies \sum a_n \text{ converges}$$

I know by the ratio test that if the limit is lower than $1$, the series will converges but what about the the superior limit?

Thanks for your help.

Answer 1

By the definition of $\limsup$ there's for each $\epsilon>0$ an $N$ such that $a_{n+1}/a_n < L+\epsilon$ whenever $n>N$. From this follows that $a_n < a_N(L+\epsilon)$. Choosing $\epsilon<1-L$ will make $a_n$ limited by a geometric series.

It's also bounded below by $0$.

Answer 2

We have $L < M < 1$ only if there is some $N \geq 1$ such that $$ \frac{a_{n+1}}{a_{n}} < M $$ for all $n \geq N$, and only if for all $p \geq 1$ we have $$ a_{N+p} < a_{N}M^{p} $$ by induction. But $\sum_{p \geq 1}M^{p}$ converges; so by comparison test $\sum_{p \geq 1}a_{N+p}$ converges.