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!

  • 3
    $\begingroup$ You cannot evaluate the limit like this, because you have $n$-terms with $n\to \infty$. $\endgroup$ Mar 12, 2019 at 15:11
  • $\begingroup$ @DietrichBurde But why? If $n\to \infty$, then it is still going to be $0+0+....=0$ infinitely $\endgroup$
    – rash
    Mar 12, 2019 at 15:13
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    $\begingroup$ @rash Certainly not. Take $n$-times $\frac{1}{n}$ as a sum. Each term goes to $0$, but the sum is always $1$. $\endgroup$ Mar 12, 2019 at 15:16
  • 1
    $\begingroup$ @rash Again how is this possible that you don't see that $\frac{1}{n}+\cdots +\frac{1}{n}=1$ for $n$ summands? $\endgroup$ Mar 12, 2019 at 15:22
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    $\begingroup$ 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 $n$. $\endgroup$
    – Mark
    Mar 12, 2019 at 15:23

1 Answer 1


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

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

  • 1
    $\begingroup$ Change "the right way" to "one right way" and I agree. See OP's question for "another way"... $\endgroup$
    – StackTD
    Mar 12, 2019 at 15:48
  • $\begingroup$ Edit made @StackTD $\endgroup$ Mar 12, 2019 at 15:50

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