A problem with root test The root test states:
If $\sum_{n=1}^{\infty}a_{n}$ is a series such that for $n\ge N$ one has $a_{n}\ge0$ then if $\lim_{n \to \infty}\sqrt[n]{a_{n}}=\rho<1$ the series $\sum_{n=N}^{\infty}a_{n}$ converges. The proof is as follows:

(I didn't mention the rest of the proof since I understand it).
The problem I have is, why do we even need to choose such epsilon? We easily can say $a_{n}<\left(\rho\right)^{n}$ for $n\ge N$; now since $\sum_{n=N}^{\infty}\left(\rho\right)^{n}$ converges (it's a geometric series with ratio $\rho<1$), using comparison test implies that the original series is also convergent. So can you tell me why we need that epsilon?
 A: I don't think that $a_n$ has to be always $< \rho^n$. I mean, if the succession $\sqrt[n]{a_n}$ was monotonically increasing (and bounded by $\rho$) in the first place, you wouldn't even need the limit! But that is not the only case where this criterion appears, it can also happen that $\sqrt[n+1]{a_{n+1}} \leq \sqrt[n]{a_{n}}$. 
It can even happen that the sequence sort of bounces arround (never getting monotone).
A: What if $\sqrt[n]{a_n}$ is always greater than $\rho$ but converges to $\rho$?
For instance, the sequence $1/n$ tends to $0$, but we cannot say that
$$
\frac{1}{n} < 0
$$
for all $n$ large.
On the other hand, for any $\epsilon>0$ there holds
$$
\frac{1}{n} < \epsilon
$$
for all $n$ sufficiently large.
A: The fact that $\sqrt[n]{a_n}\to\rho$ does not imply that eventually $\sqrt[n]{a_n}<\rho$; only that the values are as close as you want. For instance, consider $$a_n=\frac{n}{2^n}.$$ Then $\sqrt[n]{a_n}\to\frac12$, while $$\sqrt[n]{a_n}=\frac{n^{1/n}}2>\frac12$$ for all $n$. So you need the $\varepsilon$. If you are keen in avoiding it, you could choose $\gamma$ with $\rho<\gamma<1$ and then use that $\sqrt[n]{a_n}\to\rho$ implies that there exists $N$ such that $$\sqrt[n]{a_n}<\gamma$$for all $n\geq N$. But then you would be doing the same proof, just renaming $\gamma+\varepsilon$ as $\gamma$.
