# Why is $\lim\limits_{n\to\infty } 1=0$ incorrect?

$$\lim_{n\to\infty } 1 =\lim_{n\to\infty }\frac{n}{n} =\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n} =\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots =0.$$ Clearly this is incorrect, but why?

• So you're asking why $\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n} \neq \sum_{k=1}^{n} \lim_{n \to \infty} \frac{1}{n}$? Do you see how the error comes from treating $n$ as fixed in one place but as varying in the other? – Antonio Vargas Nov 24 '13 at 23:41
• @AntonioVargas: make that into an answer. I was going to answer something similar, but it feels like plagiarizing (even though it isn't really). – robjohn Nov 25 '13 at 0:16
• To add to Antonio Vargas comment, a limit $\lim_{n \to \infty}f(n)$ is an expression independent of $n$. Whereas as an expression of type $\sum_{i = 1}^{n}f(i)$ is always dependent on $n$. And this is the fundamental subtle mistake here. The rule "limit of a sum is equal to sum of limit of terms" is valid when number of terms is independent of the variable of the limit operation. Using uniform convergence the rule can be extended to handle infinite number of terms, but then also the number of terms has no relation with variable of the limit operation. – Paramanand Singh Nov 27 '13 at 5:06
• ((I guess one should add some words here about the tons of users upvoting answers that do not address the question, but sometimes, one may feel that even exercises in futility have their limits...)) – Did Nov 12 '17 at 11:58

Because $\infty\cdot0$ is undetermined. What you wrote is the same as $1=\displaystyle\lim_{n\to\infty}\frac nn=\lim_{n\to\infty}n\cdot\lim_{n\to\infty}\frac1n=$ =\infty\cdot0=\underbrace{0+0+0+...}_{\begin{align}\text{conveniently 'forgetting' to }\\\text{mention the 'number' of 0's}\end{align}}\overset{\text{"obviously"}}=0. By the same token, $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=\left(1+0\right)^\infty$ $=$ $=1\cdot1\cdot1\cdot\,...=1\neq e$.
• But that's exactly what he did. He split it into an infinite sum of $\lim_{n\to\infty}\frac1n$, all of which tend to $0$. – Lucian Nov 27 '13 at 5:17
• It seems to you that he did, but he did not express $1$ as a product of two things and use "limit of product = product of limits". He used the rule "limit of sum = sum of limits". While dealing with limits we need to very careful of expressing same thing in two different ways. – Paramanand Singh Nov 27 '13 at 6:11
• Even if we follow your point of view the real reason why this $1 = n \cdot (1/n)$ breaks is because the rule "limit of product = product of limits" is valid only when each limit exists and is finite. We should not invoke $\infty\cdot 0$ indeterminate form. If one uses the rules of limits very carefully and properly one would never get distracted by such indeterminate forms. – Paramanand Singh Nov 27 '13 at 6:13
• A product is a repeated sum. Just because something isn't mentioned explicitly doesn't mean that it's not there. Which is what you said above: The rule "limit of a sum is equal to sum of limit of terms" is valid when number of terms is independent of the variable of the limit operation. In this case it isn't, and that's what I was pointing out. I also don't understand how explicitly writing the $\infty$ contradicts what you said about the rule "limit of product = product of limits" is valid only when each limit exists and is finite. Obviously, $\infty$ isn't finite. – Lucian Nov 27 '13 at 6:29