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I just learned the definition of limits, and I learned that if $\{a_n\}, \{b_n\} $ converges, then $$\lim_{n\to \infty} (a_n+b_n)=\lim_{n \to \infty} a_n+\lim_{n \to \infty}b_n$$ holds.

And my teacher said that $\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to \infty}\frac{\frac{n(n+1)}{2}}{n^2}=\frac{1}{2}$.

But can't we compute like

$$\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\lim_{n\to\infty}\frac{2}{n^2}+\lim_{n\to\infty}\frac{3}{n^2}+\cdots +\lim_{n\to\infty}\frac{n}{n^2}=0+0+\cdots+0=0$$?

Paramanand Singh
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lminsl
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    No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl – AnyAD Oct 20 '18 at 11:44

5 Answers5

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To understand why cannot do that consider a different simpler example: $\frac 1 n +\frac 1 n+...+\frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+\cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.

Kavi Rama Murthy
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No, because what you learned was that$$\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n.$$From this, you can deduce that if you have $k$ sequences $\bigl(a(i)\bigr)_{n\in\mathbb N}$, with $i\in\{1,2,\ldots,k\}$, then$$\lim_{n\to\infty}\bigl(a(1)_n+a(2)_n+\cdots+a(k)_n\bigr)=\lim_{n\to\infty}a(1)_n+\lim_{n\to\infty}a(2)_n+\cdots+\lim_{n\to\infty}a(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.

José Carlos Santos
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No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that $$ 1 + \frac 12 + \frac 13 +\cdots = +\infty $$ while $$ 1 +\frac 1{2^2}+ \frac 1{3^2}+ \cdots = \frac {\pi^2}6 \in \Bbb R. $$ The theory of series and summation is important in calculus.

UPDATE

Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.

xbh
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    Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.” As I frequently point out, the expression $a_1+a_2+a_3+\cdots$ is not a sum — it is a limit. – MPW Oct 20 '18 at 11:54
  • @MPW Thanks for clarification. I might use the wrong vocabulary. – xbh Oct 20 '18 at 11:56
  • Well, if there is need of clarity I always prefer to state that the number of summands is finite and independent of limit variable ($n$ in question here). – Paramanand Singh Oct 20 '18 at 12:42
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As everyone mentioned, the sum rule for limits works for finite fixed number of summands.

I think using more precise notation might clarify what you've done:

$$\lim_{n\to\infty}\frac{1+2+\ldots+n}{n^2} = \lim_{\color{red}n\to\infty}\sum_{k=1}^{\Large \color{red}{n}}\frac{k}{{\color{red}{n}}^2} \stackrel{!?}= \sum_{k=1}^{\Large \color{red}{n}}\lim_{\color{red}n\to\infty}\frac{k}{{\color{red}{n}}^2} = \sum_{k=1}^{\Large \color{red}{n}} 0.$$

Basically, you left one $\color{red}n$ behind. More precisely, you simultaneously fixed $n$ and let it change to infinity. How does that work?

Really, $\lim_{n\to\infty}$ binds all occurrences of $n$; you are not allowed to move any of the $n$'s outside it's scope. If you were, it would break everything completely, for example:

$$1 = \lim_{n\to\infty} 1 = \lim_{n\to\infty}\frac nn = n \lim_{n\to\infty}\frac 1n = n\cdot 0 = 0,$$

or

$$0 = \lim_{n\to\infty} \frac 1n = \lim_{n\to\infty} \frac{n}{n^2} = \frac 1{n^2}\lim_{n\to\infty}n = \frac 1{n^2}\cdot\infty = \infty.$$

Ennar
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Moving over here from this question.

You can't, in general, convert the limit of a sum of terms into the sum of the limits of the individual terms, if the number of terms grows without bound. The problem in the original question is an example of this not working.

Recall that

$$ \lim_{n \to \infty} a_n = a $$

really means that for any $\varepsilon > 0$, there exists an integer $m > 0$ such that for all $n > m$, $| a-a_n | < \varepsilon$. The reason we can convert the limit of a sum of a fixed number of individual terms

$$ \lim_{n \to \infty} a_n + b_n + c_n + \cdots $$

into the sum of their respective limits

$$ \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n + \lim_{n \to \infty} c_n + \cdots $$

is that if there are $t$ terms, say, we can find

  • $m_a$ such that $|a-a_n| < \varepsilon/t$ whenever $n > m_a$
  • $m_b$ such that $|b-b_n| < \varepsilon/t$ whenever $n > m_b$
  • $m_c$ such that $|c-c_n| < \varepsilon/t$ whenever $n > m_c$

and so on. Then we let $m = \max\{m_a, m_b, m_c, \ldots\}$, so that

$$ |a-a_n|, |b-b_n|, |c-c_n| < \varepsilon/t $$

whenever $n > m$. Since there are $t$ terms in all, we can then conclude that

\begin{align} |(a+b+c+\cdots)-(a_n+b_n+c_n+\cdots)| & \leq |a-a_n|+|b-b_n|+|c-c_n|+\cdots \\ & < \varepsilon/t + \varepsilon/t + \varepsilon/t + \cdots \\ & = \varepsilon \end{align}

establishing the limit. But for the expression

$$ \lim_{n \to \infty} \frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n}{n^2} $$

we can't constrain ourselves to $t$ terms. Eventually the number of terms exceeds any fixed value, so that we can't divide $\varepsilon$ across the individual terms so as to obtain individual margins greater than $0$. The limit fails to be the sum of the individual limits in general (and in this case, they are different).

Brian Tung
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