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: lim

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}

=0+0=0.

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.

**Answer**

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,

**Attribution***Source : Link , Question Author : Charles Bao , Answer Author : Eric Naslund*