Showing a sequence converges to $L$ by showing that $|a_n - L| \leq K/n^p$ I would like to understand the following theorem: _"Let $(a_n:n\in \mathbb N)$ be a sequence and $L \in \mathbb R$. If "$|a_n - L| \leq K/n^p$ for some positive number $K$ and natural number $p$, for all $n \in \mathbb N$, then the sequence converges to $L$"._ However, the following proof doesn't seem to make any sense at all: _"Let $0 \lt \epsilon$ and $N \gt \sqrt[n]{K/\epsilon}$. Then $|a_n - L| \leq K/n^p \leq K/N^p \leq \epsilon$, for all $n \gt N$."_ First of all, the theorem states that it applies for _all_ natural numbers $n \in \mathbb N$. But in the proof, it only applies for numbers $n$ that are _greater_ than $\sqrt[n]{K/\epsilon}$ for each $0 \lt \epsilon$. Is this some typo I'm reading somewhere, or is there something I'm not quite understanding? Thanks in advance.

To show that $a_n \to L$ you have to show that given $\epsilon >0$ there exists $n_0$ such that $|a_n-L| <\epsilon$ for $n \geq n_0$. You don't have to prove this for all $n$; in fact, inmost cases the inequality fails for small values of $n$. In this proof $n_0=(\frac K {\epsilon})^{1/n}$.

EDIT

Of course, the hypothesis can also be weakened by saying that the inequality holds for $n \geq n_1$ for some positive integer $n_1$. But this automatically implies that the inequality holds for all $n$ with a different (larger) constant $K$. So, as far as applicability of this result is concerned we may as well say that the inequality holds for all $n$.