convergence: conditional and absolute Some sources say that the limit to infinity needs to be 0 to someone qualify for 'convergent'.
Some source say that it only needs to go to one number.
Could someone explain this?
 A: I think the confusion is in the definition of convergence. We can make the definition that a sequence of real numbers $a_n$ converges to a real number $a$ (as $n$ goes to infinity) if the sequence $a_n-a$ converges to $0$(as $n$ goes to infinity). This is currently a tautology if $a = 0$, so we say that a sequence of reals $b_n$ converges to $0$ if $\forall \epsilon > 0 \space \exists N\in \mathbb{N}$ such that for all $n \geq N, |b_n| < \epsilon$.
An equivalent definition of convergence is that a sequence of reals $a_n$ converges to $a$ if $\forall \epsilon > 0 \space \exists N\in \mathbb{N}$ such that for all $n \geq N, |a_n-a| < \epsilon$.
I personally prefer the second definition since it feels cleaner and is in a sense, "more universal", but some people prefer the first, I suppose because it's slightly more "axiomatic" in that it builds more general defitions on top of smaller ones.
A: The terms are used in different contexts. If I say:
$$
\lim_{n \to \infty} \frac{\sqrt{4 n^2 - 17}}{n} = 2
$$
what I'm saying is that the difference to 2 becomes and stays as little as I please as $n$ increases.
The other context is something like:
$$
x_{n + 1} = \frac{1}{2} \left( x_n + \frac{a}{x_n} \right) \quad x_0 = a
$$
The limit in this case happens to be $\sqrt{a}$ (See the babylonian square root method).
Such a process is said to converge if the error $x_n - \sqrt{a}$ tends to 0 when $n \to \infty$.
