I got stuck with a problem that pop up in my mind while learning limits. I am still a high school student.

Define $P(m)$ to be the statement: $\quad \lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{m})=0$

The statement holds for $m = 1$: $\quad \lim\limits_{n\to\infty}\frac{1}{n}=0$.

Assume that $P(k)$ holds for some $k$. So put $m = k$: $\quad \lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{k})=0$.

We prove $P(k + 1)$: $\quad \lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{k+1}) =\lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{k}+\frac{1}{n})$

$=\lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{k}) +\lim\limits_{n\to\infty}\frac{1}{n}$


It has now been proved by mathematical induction that statement holds for all natural m.

If we let $m=n$, then $\lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{n})=0 \tag{*}$.

However, $\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{n}=1 \implies \lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{n})=1 \tag{$\dagger$}$.

Then $(*) \, \& \, (\dagger)$ yield $1=0$?

Can anybody explain this? thanks.

  • 4
    $\begingroup$ Hint: Outside the limit, what is the meaning of the variable $n$? $\endgroup$
    – user856
    Aug 26, 2011 at 0:24
  • 81
    $\begingroup$ Let me take the opportunity to thank you for assuming that something is wrong with your argument as opposed to the nutcase assumption that you've managed to shatter the very foundations of mathematics in half a page. $\endgroup$ Aug 26, 2011 at 3:26

2 Answers 2


The problem is that the statement you proved was for fixed $m$, and then you let it vary.

What follows is one way of looking at this problem: Rewrite things as: $$\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{m}=\frac{m}{n}$$

Then what you proved by induction is that for any fixed $m$ $$\lim_{n\rightarrow \infty}\frac{m}{n}=\left(\lim_{n\rightarrow \infty}m\right)\cdot\left(\lim_{n\rightarrow \infty}\frac{1}{n}\right)=m\cdot 0=0.$$ This is fine since when the limits exist we can split them up like above. However in the second deduction you try to do the same thing $$\lim_{n\rightarrow \infty}\frac{n}{n}=\left(\lim_{n\rightarrow \infty}n\right)\cdot\left(\lim_{n\rightarrow \infty}\frac{1}{n}\right)=n\cdot 0=0.$$ This doesn't make any sense now, because $n$ is no longer fixed, and the one limit does not exist. (We only have the multiplicative property when both limits exist)

Hope that helps,

  • $\begingroup$ Uhmmm, @Eric but in your argument, I can object that $n/n=1$ except when $n=0$, so the limit should be $1$. I wanted to say, that $\lim_{n\to \infty}n/n=\infty/\infty$, but I felt uncomfortable about it. $\endgroup$
    – leo
    Aug 26, 2011 at 0:37
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    $\begingroup$ I agree with Leo. In the last statement $lim_{n\rightarrow\infty}n/n = 1$, so you also have the contradiction present in the original problem. The error is making the $n$ for $m$ substitution in the first place, not in how he took the limit. $\endgroup$
    – Clueless
    Aug 26, 2011 at 6:52
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    $\begingroup$ Of course the last line is wrong, that is the point. What I am trying to get across is that treating $n$ as if it were a fixed integer $m$ is not allowed. One way to understand this is by looking at the limits as presented above. $\endgroup$ Aug 26, 2011 at 20:26

$n$ is a free variable of the term $n$ that becomes bound during the substitution $m=n$ into


so the substitution is not logically valid.

  • $\begingroup$ A small LaTeX tip: When separating the math into a new line, I find it nicer to use $$ as environment as limits are nicer, fonts are slightly larger and it is centered. $\endgroup$
    – Asaf Karagila
    Aug 26, 2011 at 6:11
  • 1
    $\begingroup$ @Ricky : is that 'm' suppose to be actually an 'n'? +1 foe learning me free and bound variables. $\endgroup$
    – jimjim
    Aug 26, 2011 at 8:17
  • $\begingroup$ No, it's supposed to be an 'm'. The (invalid) substitution is replacing all free (which in this case is just all) instances of 'm' with 'n'. $\endgroup$
    – user57159
    Aug 26, 2011 at 10:03

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