The proof of the root test I'm reading starts with the statement: Since $ L = \lim_{n\to \infty}\sqrt[\large n]{a_n} < 1$ there exists some $L \leq r < 1$ s.t. for some $N$: $\forall n \geq N$ $\sqrt[\large n]{a_n} \leq r$ holds. But wouldn't that be false if the sequence converges from above, meaning that $\sqrt[\large n]{a_n} $ is always greater than $L$ but converges to it. Wouldn't they have to choose $r > L$ instead of $r \geq L$?
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Your comment is correct. – herb steinberg Apr 08 '21 at 16:23
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Since $L < 1,$ we know that given any $\epsilon > 0,$ for all sufficiently large values of $n$ the $n$-th roots will be within $\epsilon$ of $L.$ Thus, choosing $\epsilon = 1 - L$ (or if you don't feel like nit-picking over endpoint issues and want to be super-safe without thinking very hard, choose $\epsilon = (1-L)/2),$ it follows (draw a simple number line diagram) that for (the same) all sufficiently large values of $n$ we have the $n$-th roots less than $1.$ Note that if the $n$-th roots converge to $L$ from above, they'll still eventually have to be bounded under $1.$ – Dave L. Renfro Apr 08 '21 at 16:32
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or if you don't feel like nit-picking over endpoint issues and want to be super-safe without thinking very hard, choose --- In retrospect (and after my 5 minutes of fixing the previous comment are up), choosing $\epsilon = 1 - L$ won't quite get what we need, as having the $n$-th roots be less than $1$ for all sufficiently large values of $n$ isn't enough --- we don't want them to possibly limit at $1,$ as otherwise we can't bound things with a geometric series with ratio $r < 1.$ So we need to choose some value of $\epsilon$ that is LESS than $1-L.$ Even $\epsilon = (0.99)(1-L)$ is enough. – Dave L. Renfro Apr 08 '21 at 16:44