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**Do you have a reference to the following problem?

Show that

$$ \lim_{n\to\infty}\sum_{i=1}^n\left(1-\frac in\right)^n = \frac{1}{e-1}. $$

This problem is solved (I believe incorrectly) in

     https://www.youtube.com/watch?v=nPNB26hxLPc&t=607s

by using:

$$ \lim_{n\to\infty}\sum_{i=1}^na_{i,n}=\sum_{i=1}^{\infty}\lim_{n\to\infty}a_{i,n} $$

where $a_{i,n}$ is real.

This interchange of limit and sum is true in the given problem where $a_{i,n} = \left(1- \frac in\right)^n$but false when $a_{i,n} = \frac 1n$ .

Somos
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some person
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    Perhaps related: https://math.stackexchange.com/questions/164074/how-to-evaluate-lim-limits-n-to-infty-sum-limits-k-1n-fracknn?noredirect=1&lq=1. – StubbornAtom Jun 22 '19 at 16:06
  • Sorry, I edited the question (after the first few answers) to match the problem in the referenced website. – some person Jun 22 '19 at 16:26
  • Of course, it should be 1/(e-1). Sorry. – some person Jun 22 '19 at 17:05
  • To break down the example I think you are trying to give: $$\lim_{n\to\infty}\sum_{i=1}^n\frac in=\lim_{n\to\infty}\frac 1n\sum_{i=1}^ni=\lim_{n\to\infty}\frac 1n\frac{n(n+1)}{2}=\lim_{n\to\infty}\frac{n+1}{2}\to\infty$$ As a comparison, $$\sum_{i=1}^\infty\lim_{n\to\infty}\frac{i}{n}=\sum_{i=1}^\infty 0=0$$ – Henry Lee Jun 22 '19 at 20:33

2 Answers2

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Let's analyze when this exchange of sum and limit is allowed. It can be written as $$ \lim_{n\rightarrow\infty} \sum_{i=1}^n a_{i,n} = \lim_{m\rightarrow\infty} \sum_{i=1}^m \lim_{n\rightarrow\infty} a_{i,n}$$ We can freely add $\lim_{m\rightarrow\infty}$ in the first term and in the second term we can put $\sum_{i=1}^m$ under $\lim_{n\rightarrow\infty}$, to get the condition $$ \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=1}^n a_{i,n} = \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=1}^m a_{i,n} $$ assuming these limits are finite, we can write this condition as $$ \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \big(\sum_{i=1}^n a_{i,n} - \sum_{i=1}^m a_{i,n}\big) = 0 $$

$$ \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=m+1}^n a_{i,n} = 0$$

As you've noticed, it's not always satisfied; for $a_{i,n} =\frac1 n$ we have $$ \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=m+1}^n \frac{1}{n} = \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \frac{n-m}{n} = \lim_{m\rightarrow\infty} 1 = 1 \neq 0$$ However, if, for example, we can find a bound $|a_{i,n}| \le M_i$ such that $\sum_{i=1}^\infty M_i $ is a convergent series, then this condition is satisfied, because $$ 0 \le \left| \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=m+1}^n a_{i,n}\right| \le \lim_{m\rightarrow\infty} \lim_{n\rightarrow\infty} \sum_{i=m+1}^n M_i = \lim_{m\rightarrow\infty} \sum_{i=m+1}^\infty M_i = 0$$ In the problem from the video this condition is satisfied, because $$ a_{i,n} =(1-\frac{i}{ n})^n \le e^{-i} =: M_i$$

In conclusion a criterion that allows to get with the limit under the sum in this case is

If there exists a sequence $M_i$ such that for every $n\in\mathbb N$ we have $|a_{i,n}| \le M_i$ and $\sum_{i=1}^\infty M_i$ is convergent, then $$ \lim_{n\rightarrow\infty} \sum_{i=1}^n a_{i,n} = \sum_{i=1}^\infty \lim_{n\rightarrow\infty} a_{i,n}$$

I believe this to be a special case of Weierstrass M-test. $a_{i,n}=\frac1 n$ obviously doesn't satisfy this criterion. There may exist weaker criteria, but this one is sufficient.

Adam Latosiński
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  • I don't quite understand the first part of this, but you are basically using a form of the dominated convergence theorem. – Jair Taylor Jun 23 '19 at 17:47
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Here's how to start: $$S=\lim_{n\to\infty}\sum_{i=1}^n\left(1-\frac in\right)^n=\sum_{i=1}^\infty e^{-i}$$ this is because: $$\lim_{k\to\infty}\left(1+\frac{x}{k}\right)^k=e^x$$ now all we have left is a geometric series, which we know has the formula: $$S=a\frac{1}{1-r}=\frac{1/e}{1-1/e}=\frac{1}{e-1}$$

Henry Lee
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  • You're interchanging limit and sum which I showed by the example in my question isn't always correct. It does work in this example but needs justification. – some person Jun 22 '19 at 16:51
  • you can find a similar question here: https://math.stackexchange.com/questions/23057/under-what-condition-we-can-interchange-order-of-a-limit-and-a-summation – Henry Lee Jun 22 '19 at 16:52
  • I presume it to be true in this case because each term is positive and through the converge test the series is convergent, and the limit of the kth term $\to 0$ – Henry Lee Jun 22 '19 at 16:55
  • My counterexample satisfies all your criteria but is still gives an incorrect result. – some person Jun 22 '19 at 17:32
  • Your counter arguement is not a convergent series – Henry Lee Jun 22 '19 at 17:39
  • Lim(as n approaches inf)[Sum(i = 1 to n){i/n}] = Lim(n approarches inf)[1] =1 . Sum (i = 1 to inf)[Lim (n approaches inf) {1/n} = Sum(i = 1 to inf)[0] =0 . – some person Jun 22 '19 at 17:53
  • In your question it says 1/n – Henry Lee Jun 22 '19 at 18:07
  • You have not justified why you can interchange the limit and sum. You need to refer to a theorem that allows you to do this. – Jair Taylor Jun 22 '19 at 18:21
  • I literally linked a question detailing the same thing you are asking – Henry Lee Jun 22 '19 at 18:22
  • @HenryLee Please add that explicit justification to your answer and don't just quote the other answer. State the property which is satisfied. It is easy to overlook comments like this. – Somos Jun 22 '19 at 18:39