I can hardly imagine an easier example of the fact that my understanding of the topic is more than rusty.

I will divide the question in two parts to make the reading easier:
1) Background;
2) Problem.

In the first part, I will show how I usually approach the proof that a sequence converge to a certain limit with an example, hopefully to get if what I am doing is right or wrong. In the second part I will show my "proof" of a wrong statement, hoping to see what's wrong there for sure.


Let's assume the definition of the limit of a sequence (just to agree on the notation - with $M$ written as $M(\epsilon)$ to emphasize that it can be depend on $\epsilon$)

$$ \lim_{n \rightarrow \infty} a_n = \mathcal{l} \Longleftrightarrow \forall \epsilon >0, \exists M(\epsilon) \in \mathbb{N}: \forall m \in \mathbb{N}( m > M(\epsilon) \longrightarrow | a_m - \mathcal{l}|< \epsilon).$$

What I learned is that I should proceed in two steps, assuming the LHS of the previous definition in order to establish the RHS:
i) the "scratch work" (a.k.a. guessing-the-value-of-$M$), that does not appear in the proof,
ii) and the actual proof.


Prove that $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$.

Scratch work:
In order to find $M$, the first thing we have to notice is that

$$ \left| \frac{1}{n}-0 \right| < \epsilon \Longrightarrow \left| \frac{1}{n} \right| < \epsilon \Longrightarrow \frac{1}{n} < \epsilon \Longrightarrow \frac{1}{\epsilon} < n.$$

From this, we will set in the proof that $M= \frac{1}{\epsilon} $.

Let $\epsilon >0$ be an arbitrary real number.
Let $M= \frac{1}{\epsilon} $, from which we have that $\epsilon = \frac{1}{M}$.
Let $m$ be an arbitrary natural number, and assume that $m > M$. Thus, $\frac{1}{m} < \frac{1}{M}$. Hence,
$$ \left| \frac{1}{m} - 0 \right| = \left| \frac{1}{m} \right| = \frac{1}{m} < \frac{1}{M} = \epsilon.$$ $\square$


To state it precisely, I do have the feeling that between scratch work and proof, I can prove whatever I want, in particular false statements.


Prove that $\lim_{n \rightarrow \infty} \frac{1}{n} = 1$.

Scratch work:
In order to find $M$, the first thing we have to notice is that

$$ \left| \frac{1}{n}-1 \right| = \left| \frac{1-n}{n} \right| = \frac{1-n}{n} < \epsilon \Longrightarrow n > \frac{1}{1 + \epsilon}.$$

From this, we will set in the proof that $M= \frac{1}{\epsilon + 1} $.

Let $\epsilon >0$ be an arbitrary real number.
Let $M= \frac{1}{\epsilon + 1} $, from which we have that $\epsilon = \frac{1-M}{M}$.
Let $m$ be an arbitrary natural number, and assume that $m > M$, that is $m > \frac{1}{\epsilon + 1}$ and $\frac{1}{m} < \frac{1}{M}$.
Hence, from $m > \frac{1}{\epsilon + 1}$ we get that $\frac{1 - m}{m} < \epsilon$, that after some algebraic manipulations gives us

$$ \left| \frac{1}{m} - 1 \right| < \epsilon.$$ $\square$

Some thoughts
There is (or maybe there are) an obvious mistake that I simply cannot see. Rephrasing my problem, I feel that after that I get the value of $M$ out of the scratch work, no matter if it comes from a reasonable limit or not, it's basically done. I assume that $m > M$ and, due to the fact that this implies that $m$ has to be higher than a certain expression with $\epsilon$, I always go back to the original formula, vacously (and wrongly) "proving" the result.

Obviously it is a problem of a vicious circle, but I don't see where I start to make mistakes.
Thanks a lot for any feedback, that will be more than welcome.

  • 2
    $\begingroup$ You cannot go from $\left|\frac{1 - n}{n}\right|$ to $\frac{1 - n}{n}$, since it's negative. As $n \to \infty$, it tends to $-1$, so it's clearly less than $\epsilon$, but its absolute value isn't. $\endgroup$ – Arthur Feb 3 '14 at 11:14
  • $\begingroup$ So, summarizing, the problem is in the jump I make from the absolute value to the other expression, because it's a wrong manipulation. $\endgroup$ – Kolmin Feb 3 '14 at 12:39
  • $\begingroup$ Just as a side question. I realized that I go on a bit too automatically in trying to prove this things (a bit without having the picture and simply manipulating). Is this the standard kind of mistake you have in false proofs of these statements: automatic manipulations that hide conceptual mistakes? $\endgroup$ – Kolmin Feb 3 '14 at 12:40

You are going from

$$ \frac{1-m}{m} < \epsilon$$


$$\left| \frac{1-m}{m} \right| < \epsilon$$

Which is not true if we have:


Which is exactly what will happen if you think about it! $m$ is a natural number that will go arbitrarily high ($m>M$), so in sometime it will be greater than 1.


As an exercise, try to prove that the limit is not 1, ie, prove that for every candidate imaginable $M(\epsilon)$ there is a value $m>M(\epsilon)$ for which the "error" will be greater than $\epsilon$.


I will explain like my professor did when I was first introduced to limits.

Think of it as a game, I (the challenger) will play against you (the defendant). Your objective is to prove that the limit is a given value $L$. We play as following: I give you a number greater than zero ($\epsilon$), if you are capable to prove me that for a sufficiently large number the error is smaller than this value, you win the round. You win the match when you are capable of wining every single round.

Conversely, if I give you a single value, for which you can not give a sufficiently large number that will get you an error smaller than this value, you lose, and I win. That is, the limit is not $L$.

Let's do this for your specific case. I want you to provide me with an $M(\epsilon)$ for $\epsilon = 1/2$. But now you are in trouble! For any possible candidate $M$ you pick, I can do the following:

$$m = 2M + 1 > 2M \ge 2 \text{ (note that } m>M \text{)} $$ $$\frac{1}{m} < \frac{1}{2}$$ $$\frac{1}{m}-1 < -\frac{1}{2}<0$$ $$ \left|\frac{1}{m}-1 \right| > \frac{1}{2} = \epsilon$$

So you lose, do you see it? No matter how big, you can't get close enough to 1, so the limit isn't one. Of course that is pretty obvious if you think that you have a monotonically decreassing series $(1,1/2,1/3,1/4,1/5,\dots)$, but I was simply being extremely formal for the sake of your learning. I hope that helps elucidate things.

  • $\begingroup$ It is not clear to me where does $$\frac{1-m}{m} < - \epsilon < 0$$ come from. $\endgroup$ – Kolmin Feb 3 '14 at 12:31
  • $\begingroup$ I mean, I see where does it come from, but I am missing the whole picture here. $\endgroup$ – Kolmin Feb 3 '14 at 12:36
  • $\begingroup$ I edited my answer. Check it out. I hope it helps. $\endgroup$ – Zado Feb 4 '14 at 15:25
  • $\begingroup$ As amazing as it can sound in the XXI century, I did not have access to internet during the last days. Reading it out with the edit, thanks a lot! $\endgroup$ – Kolmin Feb 8 '14 at 7:40
  • 1
    $\begingroup$ @Kolmin, Fallacy appears at this step: $$\bigg|\frac{1-n}{n}\bigg|= \frac{n-1}{n}$$ where $n > 1.$ The LHS is positive, yet the RHS is negative. $\endgroup$ – Hoc Ngo Apr 17 '17 at 11:10

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