Proof of 1 = 0 by Mathematical Induction on Limits?

Solution 1:

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,

Solution 2:

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

$\lim\limits_{n\to\infty}(\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{m})=0$

so the substitution is not logically valid.