A basic result in analysis states that convergence of a sequence implies its boundedness. I was wondering: what's wrong with $x_n = 1/(n-a)$ for some $a \in N$? This sequence is convergent to $0$, but $x_a$ is unbounded. What am I missing here? Thanks!

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    $\begingroup$ How do you define $x_{a}$? If you plug $n=a$ you get $x_{a}=1/0$. $\endgroup$
    – Prism
    Apr 8, 2018 at 2:51
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    $\begingroup$ $x_a$ is undefined so that is not a sequence at all. $\endgroup$
    – fleablood
    Apr 8, 2018 at 3:00
  • $\begingroup$ You are missing a definition for $x_a$. $\endgroup$
    – Yanko
    Apr 8, 2018 at 16:46

5 Answers 5


The result is saying that any convergence sequence in real numbers is bounded. The sequence that you have constructed is not a sequence in real numbers, it is a sequence in extended real numbers if you take the convention that $1/0=\infty$.

  • $\begingroup$ Okay, I see. So a sequence of real numbers can be unbounded only asymptotically? $\endgroup$ Apr 8, 2018 at 3:11
  • $\begingroup$ I don't understand your question? What is asymptotically? $\endgroup$
    – user284331
    Apr 8, 2018 at 3:12
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    $\begingroup$ I mean, if you really want to say a sequence of real numbers, then you have to make sure all terms are defined. But if you take the convention that $1/0=\infty$, it is not a sequence of real numbers because $\infty$ is not in the system of real numbers. $\endgroup$
    – user284331
    Apr 8, 2018 at 3:13
  • $\begingroup$ I mean that, if you rule out all the cases like the one I asked about, an element of a sequence can be unbounded only as $n$ goes to infinity (e.g. $x_n=n$), and it's never the case that a sequence is unbounded because one of its elements $x_n$ is unbounded with $n$ finite. $\endgroup$ Apr 8, 2018 at 3:22
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    $\begingroup$ Of course, for a finite set of real numbers must be bounded, just take their maxima. $\endgroup$
    – user284331
    Apr 8, 2018 at 3:23

A real sequence is nothing but a function $$f:\mathbb{N}\longrightarrow \mathbb{R}$$ One often writes for example $x_n$ instead of $f(n)$ etc.

So, your "sequence" is not defined on $\mathbb{N}$ but on $\mathbb{N}\setminus\{a\}$.

But nicely enough, if you remove the "undefined member" by setting $x_a := r$ to an arbitrary real number $r$, you get a convergent sequence which is bounded.


Your sequence $\{ 1/(n-a)\}_1^\infty $ is not defined at $n=a$ if $a$ is a positive integer.

Thus it is not a sequence at all.

for example $$\{ 1/(n-5)\} _1^\infty = \{ -1/4,-1/3,-1/2, -1, ?, 1,...\}$$


$x_n = 1 / (n - a)$ for some natural number n is not a sequence of real numbers as all, since $x_a$ is not defined.

If you picked for example $a = 1000 - 10^{-1000}$, then $x_{1000} = 10^{1000}$. The sequence is convergent to 0, and while $x_{1000}$ is pretty large, it is still bounded by $10^{1000}$.

The simple idea behind the theorem is that the sequence will have an infinite number of elements that are not much larger than the limit, and only a finite number of elements that are not quite close to the limit. The numbers close to the limit are bounded because they are not much larger than the limit, the numbers further away from the limit are bounded because there is only a finite number of them.

There's no cheating by introducing a number to the sequence that would be unbounded on its own - that's not a real number.


I find this absurdly circular. If $x_n = 1/(n−a)$ is undefined at $n=a$, for $n$, $a$ $\in \mathbb{N}$, then the sequence is defined to not be infinite — i.e. finite and bounded — from the get go. Why then do we need a proof that $x_n$ is bounded because it converges? Any sequence is bounded given this definition.

For example, if we take $y_n = 1/x_n = (n−a)$ then, given the arguments above, $\pm \infty$ will be omitted from the range of $y_n$. Thus, $y_n$ is finite and therefore bounded.

  • $\begingroup$ You misunderstand the definition of boundedness: A sequence $(y_n)$ (let's say over the real numbers) is called bounded if there exists an $C > 0$ such that $|y_n| \leq C$. Your sequence $(y_n)$ is not bounded because for any $C > 0$ we can find some natural number $N$ which is large enough such that $|y_N| = |N - a| > C$. $\endgroup$
    – Diglett
    Sep 25, 2019 at 17:35

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