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

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

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

• 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 at 11:44

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.

• This is a very nice and easy-to-comprehend example. +1. – MPW Oct 20 at 12:26

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.

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.

• 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 at 11:54
• @MPW Thanks for clarification. I might use the wrong vocabulary. – xbh Oct 20 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 at 12:42

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

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

• In fact, the rule works if the number of summands is simply bounded, not fixed. – MPW Oct 21 at 2:20
• @MPW, actually, that's the same as fixed, as we can insert zeroes where needed. – Ennar Oct 21 at 8:21
• Yes, that’s how you would prove it. – MPW Oct 21 at 14:13

As noticed in the previous answers you can't do that because we have infinitely many summands, indeed we need to consider

$$\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to\infty} \left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right) =\lim_{n\to \infty}\sum_{k=1}^n \frac k {n^2}$$

and when you'll be aware about integration, as an alternative to the given methods, you'll be able to use Riemann sum to obtain

$$\lim_{n\to \infty}\sum_{k=1}^n \frac k {n^2}=\lim_{n\to \infty}\frac1n\sum_{k=1}^n \frac k {n}=\int_0^1 x dx=\frac12$$

• "I just learned the definition of limits" Something tells me OP doesn't know about Riemann sums yet. – Ennar Oct 20 at 13:16
• @Ennar I think you are right! I fix that, Thanks. – gimusi Oct 20 at 13:18
• What you wrote ("$\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\lim_{n\to\infty}\frac{2}{n^2}+\lim_{n\to\infty}\frac{3}{n^2}+\ldots$") is totally wrong! – user21820 Oct 20 at 13:35
• @user21820 Ops...thanks I'll take a look for that of course. Thanks to have pointed out that. – gimusi Oct 20 at 13:36
• @user21820 Thanks a lot again to have pointed out my bad bad mistake to copy and paste without checking that properly! Cheers – gimusi Oct 20 at 13:40