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Basically this: 1 = 0 with limits \begin{align} 1&=\lim_{n\to\infty} 1\\ &=\lim_{n\to\infty} \frac n n\\ &=\lim_{n\to\infty} \left(\underbrace{\frac1n + \frac1n + \cdots +\frac1n}_{n\text{ times}}\right)\\ &=\underbrace{\lim_{n\to\infty}\frac1n+\lim_{n\to\infty}\frac1n+\cdots+\lim_{n\to\infty}\frac1n}_{n\text{ times}}\\ &=\underbrace{0+0+\cdots+0}_{n\text{ times}}\\ &=0 \end{align}

Can anyone please help me find the mistake?

I know it's related to that "infinite sum" inside the brackets because it should be equal to 1, would that mean we're calculating a double limit?

Thanks!

btw, "veces" means times, and at the beginning it basically says: "what's the error"

jimjim
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Adolf RJ
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    The fourth equality is incorrect: the multiple rule $\displaystyle(\lim_{n\to\infty} kf(n)=k\lim_{n\to\infty} f(n)$ for real $k)$ doesn't apply for multiples that depend on $n.$ – ryang Jan 05 '22 at 03:17
  • the error is in the distribution of the limit to the sum of $\frac{}{n}$, you forget that you are also taking the limit as the number of summations goes to infinity – Lourenco Entrudo Jan 05 '22 at 03:20
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    Passing from the $3^{rd}$ to the $5^{th}$ line is equivalent to $\displaystyle,\lim_{n\to\infty} ,\sum_{k=1}^n\frac{1}{n} = \lim_{n\to\infty} ,\sum_{k=1}^n \lim_{n\to\infty}\frac{1}{n},$ which has no good reason to hold true, and it does not. – dxiv Jan 05 '22 at 03:35
  • I think everyone is trying to say the same thing. I would say it's like saying $\lim_{n\to\infty}n\cdot\frac 1n = \infty\cdot0=0$, which is not true as the zero in the $\infty\cdot0$ is not the absolute zero, but it's still approaching to zero, so $\infty\cdot0\ne0$. – Kay K. Jan 05 '22 at 03:38
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    Does this answer your question? Fake proof that $1=0$ using the product law for limits – Joe Jan 05 '22 at 04:24
  • Sum of the limits is not same as limit of sums, or more generally limit of a function is not function of a limit. – jimjim Jan 05 '22 at 05:21
  • @ryang Indeed, it is helpful for the OP to look at the fallacy from multiple angles. – dxiv Jan 05 '22 at 06:22

2 Answers2

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$$\lim_{n\to\infty}\frac nn \neq \lim_{n\to\infty}n\cdot\lim_{n\to\infty}\frac1n$$

Kay K.
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Naren
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  • That's why they are trivially not equal. – Naren Jan 05 '22 at 03:39
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    That is the flaw in the fake proof above, replacing "N times" in the underbraces section with multiplication by n – Alan Jan 05 '22 at 03:39
  • But it is not an equality? – Naren Jan 05 '22 at 03:43
  • @Naren Point is that the "things" on the two sides of the $\ne$ sign cannot be compared, because the RHS contains an $n$ which is nowhere defined. – dxiv Jan 05 '22 at 03:46
  • @dxiv The way you have commented is not the way he went from step 3 to 4, he didn't use a lim in from of his summation. What you showed is a more reasonable next step that he didn't take. He wrote something that doesn't parse. My answer is equally absurd but to me more obviously absurd and his was subtly absurd. A distinction that makes it more clear. – Naren Jan 05 '22 at 03:48
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    @Naren Did you intend the way I edited? – Kay K. Jan 05 '22 at 03:50
  • @dxiv Makes sense. I will keep the edit. – Naren Jan 05 '22 at 03:53
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    @Naren "he didn't use a lim in from of his summation" $;-;$ The $,\dots,$ stand for an implied limit there. – dxiv Jan 05 '22 at 03:56
  • Thanks for the tip, I see what you mean now. – Naren Jan 05 '22 at 03:58
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The fourth equality $$\lim_{n\to\infty} \left(n\left(\frac1n\right)\right) =n\left(\lim_{n\to\infty}\frac1n\right)\tag{*}$$ is incorrect:

  • the multiple rule $$\lim_{n\to\infty} kf(n)=k\lim_{n\to\infty} f(n) \quad\text{for }k\in\mathbb R$$ is inapplicable because the multiple $k$ does not depend on $n$ (so, $k$ cannot be $n$);
  • the product rule $$\lim_{n\to\infty} g(n)f(n)=\lim_{n\to\infty} g(n)\lim_{n\to\infty} f(n)$$ is also inapplicable because all the limits must exist (so, $g(n)$ cannot be $n,$ and $\displaystyle\lim_{n\to\infty}n\ne n).$

Notice also that $(*)$ is equivalent to $$\lim_{m\to\infty} \left(m\left(\frac1m\right)\right) =n\left(\lim_{m\to\infty}\frac1m\right),$$ and that $$\exists n\;\lim_{m\to\infty} \left(m\left(\frac1m\right)\right) =n\left(\lim_{m\to\infty}\frac1m\right)$$ is a false statement.

ryang
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