limit $\underbrace{\lim_{n\to\infty}\frac1n+\lim_{n\to\infty}\frac1n+\cdots+\lim_{n\to\infty}\frac1n}_{n\text{ times}}$

Question

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"

Answer 1

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

Answer 2

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.