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

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    $\begingroup$ 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? $\endgroup$ – Antonio Vargas Nov 24 '13 at 23:41
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    $\begingroup$ @AntonioVargas: make that into an answer. I was going to answer something similar, but it feels like plagiarizing (even though it isn't really). $\endgroup$ – robjohn Nov 25 '13 at 0:16
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Nov 27 '13 at 5:06
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    $\begingroup$ @1110101001 Interestingly, you declare that you "still don't understand why (etc.)" one month after having accepted an answer. So, in the end, do you consider that the answer you accepted addresses your problem, or not? Please be reminded that on this site, accepting an answer is declaring ipso facto that it does address one's question, hence there seems to be a clear logical contradiction here. (In addition, I may mention my own analysis, which is that the accepted answer does not address your problem, but, in a first phase, this opinion (shared by others) is even irrelevant.) $\endgroup$ – Did Nov 12 '17 at 11:54
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    $\begingroup$ ((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...)) $\endgroup$ – Did Nov 12 '17 at 11:58
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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, $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=\left(1+0\right)^\infty$ $=$ $=1\cdot1\cdot1\cdot\,...=1\neq e$.

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    $\begingroup$ OP has not split the limit into the way you have mentioned in your answer. I don't see how your point of view answers his question. Please check the comments from Antonio Vargas. He has mentioned the real issue with OP's approach. $\endgroup$ – Paramanand Singh Nov 27 '13 at 5:04
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    $\begingroup$ 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 $0$. $\endgroup$ – Lucian Nov 27 '13 at 5:17
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Nov 27 '13 at 6:11
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Nov 27 '13 at 6:13
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    $\begingroup$ 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. $\endgroup$ – Lucian Nov 27 '13 at 6:29
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Unlike what Lucian said, it is possible to give a precise meaning (think measure theory) to an infinite sum of limits, and the answer can be 0 under suitable conventions. However, the problem was in the step of splitting the limit. There is a theorem that shows that you can split a limit of a sum of two terms into the sum of their limits provided they exist. That can never result in an infinite sum.

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  • $\begingroup$ I still don't understand why you can't split a sum of terms when the number of terms is dependent upon the variable of the limit. Could you explain further why this can't be? $\endgroup$ – 1110101001 Dec 29 '13 at 2:28
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    $\begingroup$ Do you understand the proof of the theorem that the sum of two limits is the limit of their sum? If you do, you will see that you cannot prove the theorem for an infinite sum of limits. All you can prove (by induction) is for a finite sum of limits. $\endgroup$ – user21820 Dec 29 '13 at 2:30

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