0
$\begingroup$

Why this definition is not a good definition?
"As $n \to \infty$, $a_n$ gets closer and closer to $L$, $L$ is the limit"

(a) Can a sequence satisfy this definition and still fail to converge?
(b) Can a sequence converge yet fail to satisfy this definition?

  • 6
    Well....let $a_n=\frac 1n$. Then it is certainly true that as $n\to \infty$ we have "$a_n$ gets closer to $-1$" , yes? But $-1$ is not the limit. – 2017-01-22
  • 3
    How do you define _closer and closer_ .? – 2017-01-22
  • 0
    $a_n = {1\over n}$ still converges to 0 though – 2017-01-22
  • 3
    not following. Your definition doesn't define a limit because, even ignoring the imprecise nature of "closer and closer" there may be multiple values of $L$ that $a_n$ grows closer to. Or, as in the example, infinitely many values. – 2017-01-22
  • 1
    Let $L=-1$ and $a_n = {1 \over n}$, then as $n \to \infty$, $a_n$ gets closer and closer to $L$, however, clearly the limit is zero. – 2017-01-22
  • 1
    Ok: define a sequence $a_n$ by: $a_{2n}=\frac 12 +\frac 1n$ and $a_{2n+1}=-\frac 12 - \frac 1n$. In this case $a_n$ grows closer to $0$ as $n$ grows, but the sequence does not converge. – 2017-01-22
  • 0
    Apart from being a "bad" definition it is also misleading. "Closer and closer" does not capture the idea. For example $a_{2n}=2^{-n}$ and $a_{2n+1} = 2^{-n} + n^{-1}$ is a sequence that does not get "closer and closer" to $0$ but we want a definition that does allow us to say that $\lim a_n=0$. The usual loose version in general use replaces "closer and closer to $L$" with the equally vague (but better) "arbitrarily close to $L$." – 2017-01-23

1 Answers 1

1

It's too vague to be a real definition. What exactly "$a_n$ gets closer to $L$" means is vague. The proper def'n is: $$\lim_{n\to \infty}a_n=L\iff \forall r>0\;\exists m\;(\;\forall n>m\;(|a_n-L|

Another, equivalent, def'n is that for any $r>0$ the set $\{n: |a_n-L|\geq r\}$ is finite.

  • 0
    Might be worth adding that even if we accept "closer and closer" as a rigorous term (which it obviously isn't), then there are infinitely many values to which a series may get "closer and closer" without that actually being the limit according to the conventional (i.e. your) definition. The series $a_n = \frac{1}{n}$ has been mentioned above, which obviously gets "closer and closer" to every single negative number as $n$ goes to infinity, even though the proper limit is $0$. – 2017-01-22
  • 1
    @Kurow .Well said. And "closer and closer" may give the (incorrect) impression that convergence must be monotonic. – 2017-01-24