Conditions for convergence to 0 of 1/n^a Apologies if this has already been asked and answered, I could not find anything explicit on it.
I'm trying to find the range of $\alpha$ such that $$ \lim_{n\to\infty} \frac{k}{n^\alpha} = 0 $$
with $k>0$.
Now obviously $ \alpha \ge 1 $ is fine, but intuitively I believe $\alpha > 0$ is sufficient. Is this correct and if so, why?
I have considered thinking of it as a sequence so
$$x_n = \frac{k}{n^\alpha}$$
Then for $ M>n$, 
$$ \vert x_M - L \vert <\epsilon $$
$$ \therefore \vert \frac{k}{M^\alpha}-0 \vert < \epsilon $$
$$ \therefore \frac{k}{M^\alpha} < \epsilon $$
But I can get no closer to a definitive answer.
Thanks in advance to whomever can help here!
 A: Suppose $\alpha > 0$, then let $\epsilon > 0$. Note that: $\frac{k}{n^\alpha} = \exp (\log \frac{k}{n^\alpha}) = \exp(\log k - \alpha \log n)$.
Now, if $n$ increases to infinity, then $\log n$ also increases to infinity. So the whole thing hinges on $\alpha$.
If $\alpha$ is positive, then $\log k - \alpha \log n$ will keep decreasing, then it will go to $- \infty$ (roughly, you need to make this rigorous), and $e^{-\infty} = 0$ (once again, roughly), therefore the limit is zero.
On the other hand, if $\alpha$ is negative, then $\log k - \alpha \log n$ will keep increasing, so that it goes to $+\infty$, and $e^{+\infty} = +\infty$, so convergence doesn't happen.
Of course, if $\alpha = 0$, then the limit is $k$.
This answer is not rigorous, however it wishes to point out the power of rewriting the given expression in a form where it is almost immediately obvious  that the values of $\alpha$ convergence occurs are only the non-negative ones.
(By roughly, I mean to say that the understanding of the statement should be clear, along with a rigorous proof, but the statement itself is purely indicative, and in fact incorrect as a mathematical statement).
A: Let $f(x)=k/x^\alpha.$ Note that if we put $u=1/x$ then as $x \to +\infty,$ we have $u \to 0^+,$ and that in terms of $u$ we have $f=ku^\alpha.$ Since for $\alpha>0$ the latter function is continuous from the right at $0,$ your statement is right about the limit of $f$ at infinity.
Some of this conclusion depends on verifying the limits may be substituted in as suggested, and in showing $x^\alpha$ continuous from the right at $0.$
