Proof of 1 = 0 by Mathematical Induction on Limits?

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

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