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$$ \lim_{x\to \infty}\left(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} + ... + \frac{x}{x^2}\right) = \lim_{x\to \infty} \left(\frac{x(x+1)}{2x^2}\right) $$ $$ = \lim_{x\to \infty} \left(\frac{(1+\frac{1}{x})}{2}\right) $$ $$ = \frac{1}{2}$$

However,

$$ \lim_{x\to \infty}\frac{1}{x^2} + \lim_{x\to \infty}\frac{2}{x^2} + \lim_{x\to \infty}\frac{3}{x^2} + ... + \lim_{x\to \infty}\frac{x}{x^2} = 0$$

Where am I going wrong with this?

Akshit Chhabra
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  • $\lim_{x\rightarrow\infty} \frac{1+\frac{1}{x}}{x} = 0$. – stange Jun 17 '23 at 18:05
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    There are differing number of terms in the sum. If you were to write $\frac{x}x=\frac1x+\cdots+\frac1x,$ you'd have a much simpler form of the paradox. – Thomas Andrews Jun 17 '23 at 18:06
  • @stange that was a typing error. I have edited the questions and fixed it. – Akshit Chhabra Jun 17 '23 at 18:09
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    @ThomasAndrews yes but why does this happen... All limits are individually defined so we should be able to write limit of sum as sum of limits... – Akshit Chhabra Jun 17 '23 at 18:10
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    Also, there is a tricky notation thing here. $\lim_{x\to \infry}$ and $\lim_{n\to\infty}$ mean slightly different things - the first means $x$ is a real number, the second restricts to integer $n.$ This is never taught, you just have to learn it. Your formula only makes sense when $x$ is an integer, so you should really switch to an integer-signaling variable, $n$ or $m$ or $k,$ rather than a real-signaling variable. – Thomas Andrews Jun 17 '23 at 18:11
  • It is true that if $f(n,k)$ is a function then: $$\lim_{n\to\infty}\sum_{k=1}^K f(n,k)=\sum_{k=1}^{K} \lim_{n\to\infty} f(n,k),$$ when the limits on the right side converge, But your $K$ varies per $n.$ It is not true that: $$\lim_{n\to\infty} \sum_{k=1}^{\infty} f(n,k)=\sum_{k=1}^{\infty}\lim_{n\to\infty} f(n,k)$$ – Thomas Andrews Jun 17 '23 at 18:17
  • Seconding @ThomasAndrews' comment: it's not any sort of "mathematical law", but notation does "signal" things: currently, "$x$" is probably a real variable, while "$n$" is probably an integer variable. And, in your story, the expression really only makes sense for an integer $x$, so it'd be less cognitively-dissonant for your readers to call it $n$. It's "signalling", yes, not formal correctness, but for human readers that does matter a lot. – paul garrett Jun 17 '23 at 18:18
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    The inability to say, in general, that the limit of an infinite sum is the infinity sum of the limits of the terms is the reason we need theorems like the Dominated Convergence theorem and the Monotone Convergence Theorem, both of which gives specific cases where we can do so. – Thomas Andrews Jun 17 '23 at 18:34
  • And while this is technically a finite sum for each $x,$ we can't treat it a finite sum of a particular number of terms, because the number of terms is $x$ for each sum, and this is not bounded. – Thomas Andrews Jun 17 '23 at 18:35
  • An interesting additional case is: $$f(n,k)=\begin{cases}1&n=k\0&n\neq k\end{cases}.$$

    Then for each $k,$ $$\lim_{n\to\infty} f(n,k)=0,$$ but $1=\sum_{k=1}^\infty f(n,k)$ for any $n.$

     – Thomas Andrews Jun 17 '23 at 20:06
  • Maybe it is because $\sum_{i=1}^{\infty} \lim_{x \to \infty} \frac i x$ is already the infinite sum. – RDK Jun 18 '23 at 15:59
  • Also relevant: https://math.stackexchange.com/questions/1598719/, https://math.stackexchange.com/questions/2242859/, https://math.stackexchange.com/questions/2963178/. – Jam Jun 18 '23 at 18:52
  • I think the terse answer by user137731 on Question 1598719 answers your question best: "... the number of terms goes up exactly as the size of each term goes down". – Jam Jun 18 '23 at 18:57

1 Answers1

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As noticed, the limit in this form

$$\lim_{x\to \infty}\overbrace{\left(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} + ... + \frac{x}{x^2}\right)}^{\text{x terms}}$$

is meaningful for $x$ integer and in this case, as you noticed

$$\lim_{x\to \infty}\left(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} + ... + \frac{x}{x^2}\right) = \lim_{x\to \infty} \left(\frac{x(x+1)}{2x^2}\right)=\frac12$$

note also that the limit in the last form

$$\lim_{x\to \infty} \left(\frac{x(x+1)}{2x^2}\right)=\frac12$$

is well defined also for $x$ real.

The apparent paradox you are referring to

$$\lim_{x\to \infty}\left(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} + ... + \frac{x}{x^2}\right)=\lim_{x\to \infty}\frac{1}{x^2} + \lim_{x\to \infty}\frac{2}{x^2} + \lim_{x\to \infty}\frac{3}{x^2} + ... + \lim_{x\to \infty}\frac{x}{x^2} = 0$$

is due to the fact that the following rule

$$\lim_{x\to x_0} f_1(x)\cdot f_2(x)\cdot \ldots\cdot f_n(x)=\lim_{x\to x_0} f_1(x)\cdot \lim_{x\to x_0} f_2(x)\cdot \ldots\cdot \lim_{x\to x_0} f_n(x) $$

holds, under some conditions for the existence of the single limits, only for $n$ finite otherwise we face with apparent paradoxes as the one in hand.

Note also that this wrong argument in some cases works, as for example for

$$\lim_{x\to \infty}\overbrace{\left(\frac{1}{x^3} + \frac{2}{x^3} + \frac{3}{x^3} + ... + \frac{x}{x^3}\right)}^{\text{x terms}}= 0$$

but notwithstanding this, it is not correct claim that

$$\lim_{x\to \infty}\left(\frac{1}{x^3} + \frac{2}{x^3} + \frac{3}{x^3} + ... + \frac{x}{x^3}\right)=\lim_{x\to \infty}\frac{1}{x^3} + \lim_{x\to \infty}\frac{2}{x^3} + \lim_{x\to \infty}\frac{3}{x^3} + ... + \lim_{x\to \infty}\frac{x}{x^3}$$

even if it leads to a correct result in this particular case.

user
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