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

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:

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

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:
$$limsup_{nto infty} frac{a_{n+1}}{a_n}=R$$
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 $kgeq N$:
$${a_k} leq a_N(R+ epsilon)^{k-N}.$$
and this can be rewritten to:
$$ {a_k} leq a_N(R+ epsilon)^k (R+ epsilon)^{frac{1}{N}} .$$
$$ (a_k)^{frac{1}{k}} leq a_N ^{frac{1}{k}}(R+ epsilon) (R+ epsilon)^{frac{-N}{k}} .$$

If I take the limit and supremum for $kto infty$ 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 $2epsilon$?):
$$limsup_{kto infty}(a_k)^{frac{1}{k}} leq R + 2epsilon$$
Which can be used to conclude that (how did they just drop the $epsilon$?):
$$limsup_{kto infty}(a_k)^{frac{1}{k}} leq R. $$

Which is equivalent to the desired equality between the limsups.