# Series, computing the limit $\lim\limits_{n\to\infty}\frac{1+2+\cdots+n}{n^3}$

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...

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

## 3 Answers

Just for fun:

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

• Did you just edit that?? (I'm feeling like I was seeing things...) – apnorton Aug 2 '13 at 15:03
• What did you see? – OR. Aug 2 '13 at 15:06
• Thanks, could I ask you for the explanation of the part containing the integral? – justik Aug 2 '13 at 15:11
• @RGB I saw $\lim \cdots = 1$ – apnorton Aug 2 '13 at 15:12
• @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. – OR. Aug 2 '13 at 15:16

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

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

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

and this goes to zero.

• Corrected the error! – Vishal Gupta Aug 3 '13 at 3:52