Can't we compute $\lim_{n \to \infty}\frac{1+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\cdots +\lim_{n\to\infty}\frac{n}{n^2}=0$?

Question

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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

Answer 5

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).