7

Evaluate $$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3}$$

This is the original method to solve this is:

Taking summation of the square numbers $1^2 + 2^2 + 3^2 +...+n^2 = \frac{1}{6}n(n+1)(2n+1)$ $$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \frac{\frac{1}{6}n(n+1)(2n+1)}{n^3}$$ $$=\lim\limits_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim\limits_{n\to \infty} \frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})$$ $$=\frac{2}{6} = \frac{1}{3}$$

But when looking at the limit in a different angle I get a different answer,

$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \lim\limits_{n\to \infty} \frac{1^2}{n^3}+\frac{2^2}{n^3}+...+\frac{1}{n}$$ $$=0+0+...+0 = 0$$

Both the method seem right to me, but why I am getting different answers? What have I done wrong? Please Explain. Thank you!

rash
  • 2,070
  • 3
    You cannot evaluate the limit like this, because you have $n$-terms with $n\to \infty$. – Dietrich Burde Mar 12 '19 at 15:11
  • @DietrichBurde But why? If $n\to \infty$, then it is still going to be $0+0+....=0$ infinitely – rash Mar 12 '19 at 15:13
  • Your claim in the 2nd part is not true, as for any $\epsilon > 0$ you can not find any $K\in\mathbb N$ for which $|\frac{1^2+2^2+\dots +n^2}{n^3}|<\epsilon$ for all $n>\ge K$ – Sujit Bhattacharyya Mar 12 '19 at 15:14
  • 4
    @rash Certainly not. Take $n$-times $\frac{1}{n}$ as a sum. Each term goes to $0$, but the sum is always $1$. – Dietrich Burde Mar 12 '19 at 15:16
  • @DietrichBurde Again how is it possible? Sorry, I do not get u well... – rash Mar 12 '19 at 15:18
  • Because, as you showed in your first answer, when you add an infinite number of terms, each of which tends to zero, the sum can be nonzero. An easy example of this is n terms each being 1/n. The sum is 1 while each term tends to zero. – marty cohen Mar 12 '19 at 15:18
  • @rash: What is important is that you have $n$ terms. That is, try to look at it this way: $c_n=a_1+a_2+\cdots+a_n$. It doesn't make sense to say $$\lim_{n\to\infty}c_n=\left(\lim_{n\to\infty}a_1\right)+\cdots\left(\lim_{n\to\infty}a_n\right)$$ since as $n$ becomes large, $c_n$ includes more and more terms whereas on the right, you're acting as though the number of terms is fixed. – Clayton Mar 12 '19 at 15:19
  • 1
    @rash Again how is this possible that you don't see that $\frac{1}{n}+\cdots +\frac{1}{n}=1$ for $n$ summands? – Dietrich Burde Mar 12 '19 at 15:22
  • 3
    Arithmetic of limits works when the number of sequences in the sum is finite. But you can't use it for a number of sequences that is dependent on $n$. – Mark Mar 12 '19 at 15:23
  • Possible duplicate of this question? – Robin Sep 16 '20 at 14:21

1 Answers1

3

Converting the limit into an integral is one right way to evaluate it.

$$\lim_{n\to \infty}\dfrac{1^2+2^2+\cdots+n^2}{n^3}=\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^2=\int_{0}^{1}x^2\mathrm dx$$

Paras Khosla
  • 6,439