0
$\begingroup$

How to compute the following limit? The series is given by

$$\lim_{n\rightarrow\infty}\frac{1}{n^3}(1+2+\cdots+n)$$

Thanks for your help...

$\endgroup$
  • 2
    $\begingroup$ Do you know a formula for the sum $1+2+\cdots+n$? $\endgroup$ – David Mitra Aug 2 '13 at 14:54
  • 6
    $\begingroup$ Hint: $1+2+\cdots+n\leqslant n+n+\cdots+n=n^2$. $\endgroup$ – Did Aug 2 '13 at 14:56
  • $\begingroup$ This is not a series, it is a sequence. $\endgroup$ – Thomas Andrews Aug 2 '13 at 15:47
4
$\begingroup$

Just for fun:

$$\lim_{n\rightarrow\infty}\frac{1}{n}\left(\frac{1}{n}\frac{1}{n}+\frac{2}{n}\frac{1}{n}+\ldots+\frac{n}{n}\frac{1}{n}\right)=\lim_{n\rightarrow\infty}\frac{1}{n}\times\int_{0}^{1}\text{d}x=0$$

$\endgroup$
  • $\begingroup$ Did you just edit that?? (I'm feeling like I was seeing things...) $\endgroup$ – apnorton Aug 2 '13 at 15:03
  • $\begingroup$ What did you see? $\endgroup$ – OR. Aug 2 '13 at 15:06
  • $\begingroup$ Thanks, could I ask you for the explanation of the part containing the integral? $\endgroup$ – justik Aug 2 '13 at 15:11
  • $\begingroup$ @RGB I saw $\lim \cdots = 1$ $\endgroup$ – apnorton Aug 2 '13 at 15:12
  • 1
    $\begingroup$ @justik It is a joke. Just a convoluted way of proving it. If you take the definition of the integral of the constant function $1$ using Riemann sums and take uniform partitions of $n$ steps you get what is inside the brackets. $\endgroup$ – OR. Aug 2 '13 at 15:16
2
$\begingroup$

$\lim_{n\to\infty}\frac{1+2+...+n}{n^{3}}=\lim_{n\to\infty}\frac{\frac{n(n+1)}{2}}{n^{3}}=\lim_{n\to\infty}\frac{1}{2n}\big(1+\frac{1}{n}{}\big)=0$

$\endgroup$
-2
$\begingroup$

$1 + \cdots + n = n(n+1)/2$ and hence

$$ \frac{1}{n^{3}}(n(n+1))/2 = \frac{1}{2n} + \frac{1}{2n^{2}} $$

and this goes to zero.

$\endgroup$
  • 1
    $\begingroup$ Corrected the error! $\endgroup$ – Vishal Gupta Aug 3 '13 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.