$ \limsup_{k\to \infty} (a_k)^{1/k} \leq \limsup_{k\to \infty} \frac{a_{k+1}}{a_k} $

Question

I am working on the following problem, let $(a_n)_{n \in \mathbb N}$ be a sequence of positive real numbers, we wish to show that:

$$ \limsup_{k\to \infty} (a_k)^{1/k} \leq \limsup_{k\to \infty} \frac{a_{k+1}}{a_k} $$

a similar problem as in (but which does not answer my own questions/confusions):

Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$

We have: $$\limsup_{n\to \infty} \frac{a_{n+1}}{a_n}=R$$ I understand that from the limit definition we have for any $\epsilon>0$ there exists an $N$ such that $n>N$ means: $$\frac{a_{n+1}}{a_n} \leq R+ \epsilon.$$ We can then arrive at the following inequality using the one above for $k\geq N$: $${a_k} \leq a_N(R+ \epsilon)^{k-N}.$$ and this can be rewritten to: $$ {a_k} \leq a_N(R+ \epsilon)^k (R+ \epsilon)^{-N} .$$ $$ (a_k)^{\frac{1}{k}} \leq a_N ^{\frac{1}{k}}(R+ \epsilon) (R+ \epsilon)^{\frac{-N}{k}} .$$

If I take the limit and supremum for $k\to \infty$ I find ($1 \cdot (R+\epsilon)\cdot 1$): $$ \limsup_{k \to \infty } (a_k)^{\frac{1}{k}} \leq R+\epsilon $$ The question tells me to derive that following inequality (why $2\epsilon$?): $$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R + 2\epsilon$$ Which can be used to conclude that (how did they just drop the $\epsilon$?): $$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R. $$

Which is equivalent to the desired equality between the limsups.

Answer 1

You have a typo, there is a $1/N$ exponent that should be $-N$.

The $2\epsilon$, you don't need it the way you wrote the answer.

As for "how did they just drop the $\epsilon$", if $a\leq R+\epsilon$ for all $\epsilon>0$, then $a\leq R$. This is simply the fact that $a-R$ is less than every positive number, so it has to be less than or equal zero.

Answer 2

1. Forget about the $2\epsilon$ thing. Take upper limits (limsup) and you get $ \limsup_{k \to \infty } (a_k)^{\frac{1}{k}} \leq R+\epsilon$. That's perfect.

2. We can drop the $\epsilon$ because we chose $\epsilon$ arbitrarily. This is a standard technique. For example, if one has difficulty understanding $5 \leq 5$, we may convince him by showing $5 < 5 + 1, 5 < 5 + 0.2, 5 < 5 + 0.0002 \dots$, and finally $5 < 5 + \epsilon$ for arbitrary $\epsilon$.