2

Does anyone know why bringing the limit inside the sum doesn't work sometimes, other than the fact that sometimes the resulting answer contradicts the answer when the limit is not inside the sum?

Math12345
  • 321
  • 2
    I believe you should first take a look at your last question. It's already answered. – The Dead Legend Apr 23 '17 at 01:01
  • 2
    @dxiv It appears the OP is not satisfied with the counterexamples given in response to his last question. Instead, he would now like an answer that shows why one wouldn't generally expect it to be possible to move the limit inside a sum. He has added this clarification to his new question. – user49640 Apr 23 '17 at 01:06
  • @user49640 The OP essentially replaced the "basic way to understand" from the first question with "conceptual understanding" here. I don't see how that changes the question in any substantive way. Requests for clarification belong in the original question, not in duplicate posts. – dxiv Apr 23 '17 at 01:15
  • @dxiv The OP has added that he doesn't just want a counterexample. – user49640 Apr 23 '17 at 01:18
  • @user49640 Right, but that doesn't change the question, and should have been edited into the original question, instead. – dxiv Apr 23 '17 at 01:20
  • 1
    @dxiv I'm not sure, I thought when you realize you've asked your question in the wrong way, and you've already received good answers to the wrong question, you don't change your question. – user49640 Apr 23 '17 at 01:22

2 Answers2

2

I think that it's easier to understand this conceptually if you consider exchanging integrals and limits. After all integration is a sort of "general summation."

That said, we are now asking the question of when do $\lim$ and $\int$ commute? That is, given a sequence of functions $\{f_n\}$ such that $f_n\to f$, when is it the case that $$ \lim_{n\to\infty}\int f_n = \int \lim_{n\to\infty} f_n? $$ Well, a first course in real analysis says that if the convergence of $f_n \to f$ is "regular enough" or uniform enough, then these operations should trade places. To see an example of when it doesn't work, consider the sequence of functions defined by $$ f_n(x) = \begin{cases} n^2x & \text{if $0\le x\le \frac{1}{n}$} \\ 2n-n^2x & \text{if $\frac{1}{n}\le x\le \frac{2}{n}$} \\ 0 & \text{if $\frac{2}{n}\le x\le 1$.} \\ \end{cases} $$ If we plotted these for a few values of $n$, we would see that these look a series of "tents" that are growing and sliding over to the origin, but their mass is never actually changing.

Computing the integral of $f_n$ gives $1$ if I wrote down $f_n$ correctly, so $\lim_{n\to\infty}\int f_n = 1$. On the other hand, for any given value of $x$, eventually our tent slides past it, so that $\lim_{n\to\infty}f_n(x) = 0$. Hence, when we integrate the limit function, we are integrating $0$, and of course $\int \lim_{n\to\infty}f_n = \int 0 = 0$.

On the other hand, if the sequence of functions we considered were very nicely behaved, such as $f_n(x) = \frac{1}{n}$, then we see that the integral and the limit do commute: $$ \lim_{n\to\infty}\int \frac{1}{n} = \lim_{n\to\infty}\frac{1}{n}= 0 \quad\text{and}\quad \int\lim_{n\to\infty}\frac{1}{n} = \int 0 = 0. $$ Here of course, the mass of the function is moving in the same direction as the direction in which the function is converging, so to speak. So the intuitive slogan is, if the mass moves in the general direction of convergence, we should expect to trade operations fluidly.

Alex Ortiz
  • 24,844
1

Let's turn the question around: Why should the limit of a sum be the sum of the limits?

For finite sums, of course, there's no issue. "Infinite sums", by contrast, entail a second limit operation, e.g., $$ f(x) = \sum_{k=0}^{\infty} f_{k}(x) = \lim_{n \to \infty} \sum_{k=0}^{n} f_{k}(x). $$ To say "the limit of a sum is the sum of the limits" is to interchange two limit operations: $$ \lim_{x \to x_{0}} f(x) = \sum_{k=0}^{\infty} \lim_{x \to x_{0}} f_{k}(x) $$ if and only if $$ \lim_{x \to x_{0}} \lim_{n \to \infty} \sum_{k=0}^{n} f_{k}(x) = \lim_{n \to \infty} \lim_{x \to x_{0}} \sum_{k=0}^{n} f_{k}(x). $$ We can now see that taking a sum is a gratuitous complication: A sequence of partial sums is merely a (different) sequence. Consequently, we may as well ask whether $$ \lim_{x \to x_{0}} \lim_{n \to \infty} F_{n}(x) = \lim_{n \to \infty} \lim_{x \to x_{0}} F_{n}(x) $$ for an arbitrary sequence of functions, i.e., for a function of one natural number and one real or complex variable; or for a double sequence, i.e., a real- or complex-valued function of two natural numbers.

The plain fact is, iterated limits need not commute, even if both limits exist. Illustrative examples include $$ 1 = \lim_{n \to \infty} \lim_{x \to 0} (1 + x^{2})^{-n} \neq \lim_{x \to 0} \lim_{n \to \infty} (1 + x^{2})^{-n} = 0, $$ or $$ 1 = \lim_{n \to \infty} \lim_{m \to \infty} \frac{m}{m + n} \neq \lim_{m \to \infty} \lim_{n \to \infty} \frac{m}{m + n} = 0, $$ or, with $a \neq b$ distinct (but otherwise arbitrary) real numbers, $$ a = \lim_{n \to \infty} \lim_{m \to \infty} \frac{am + bn}{m + n} \neq \lim_{m \to \infty} \lim_{n \to \infty} \frac{am + bn}{m + n} = b. $$