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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?

user68610
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    Hi, welcome to M.SE. Try being a bit more specific when asking questions so it is clear what exactly you are talking about. What sources, what limit? – muzzlator Mar 26 '13 at 17:40
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    @user68610 Can you provide some context? You must be confusing something.The limit need only be a real number for a real sequence to converge. – Git Gud Mar 26 '13 at 17:40

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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.

Tom Oldfield
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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 $n \to \infty$.

vonbrand
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