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):
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.
Best Answer
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.