Finding the Limit of $\frac{1}{n^2} [n + (n+1) + ...+ 2n]$ I was studying for some quizzes when I encountered this problem. It goes like this:
Find the limit of $\frac{1}{n^2} [n + (n+1) + ...+ 2n]$
My work:
I do know that the pattern looks like this:
0
1+2
2+3+4
3+4+5+6
4+5+6+7+8
5+6+7+8+9+10
(note the first and the last terms)
and the list goes on....
But I don't know what is the equation that describe that sequence above.
If I know the sequence, I can able to solve for the limit like this:
$$ \lim_{n\to \infty}  \frac{1}{n^2} (the \space sequence \space that \space describe \space the \space pattern \space above)$$
How to get the sequence, and ultimately, the limit of the sequence described above?
 A: You can rewrite the general term of your sequence more compactly as $\frac{1}{n^2}\sum_{k=0}^n (n+k)$. 
By recognizing a Riemann sum, you then will be able to obtain the limit by computing the corresponding integral:
$$
\frac{1}{n^2}\sum_{k=0}^n (n+k)
= \frac{1}{n}\sum_{k=0}^n (1+\frac{k}{n}) 
\xrightarrow[n\to\infty]{} \int_0^1 (1+x)dx = \frac{3}{2}
$$

Note that this is simply one way to compute the limit (which has the advantage to generalize to other similar questions -- Riemann sums are a nice arrow to have in your quiver): for instance, you could also use the fact that $$\sum_{k=0}^n (n+k) = \sum_{k=0}^n n + \sum_{k=0}^n k = n(n+1) + \frac{n(n+1)}{2} = \frac{3}{2}n(n+1)$$ and then remark that $\lim_{n\to\infty}\frac{n(n+1)}{n^2} = 1$.
A: Since$$1+2+3\cdots+2n=\frac{2n(2n+1)}2$$and$$1+2+\cdots+(n-1)=\frac{(n-1)n}2,$$you have\begin{align}n+(n+1)+\cdots+2n&=\frac{2n(2n+1)}2-\frac{(n-1)n}2\\&=\frac{3n^2+3n}2.\end{align}Therefore,$$\lim_{n\to\infty}\frac1{n^2}\bigl(n+(n+1)+\cdots+2n\bigr)=\lim_{n\to\infty}\frac{3+\frac3n}2=\frac32.$$
A: $$\begin{align}\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=0}^{n}(n+k)&=\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=1}^{n}(n+k)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\Big(1+\frac{k}{n}\Big)\\&=\int_{0}^{1}(1+x)dx=\Big[x+\frac{x^2}{2}\Big]=\frac{3}{2}\end{align}$$
A: The sum of an arithmetic progression is the number of terms times the average of the extreme terms. Hence the limit is $$\frac32.$$
A: $$[n+0]+[n+1]+[n+2]+...+[n+n]=[0+1+2+...+n]+(n+1)\cdot n=\\=\frac{n(n+1)}{2}+n(n+1)=1.5n(n+1)=1.5n^2+1.5n\\\\\lim_{n\to \infty}\frac{1.5n^2+1.5n}{n^2}=1.5$$
A: The square brackets
$$[n + (n+1) + …+ 2n]
$$
represent an arithmetic progression whose sum is the average of the first and last terms, times the number of terms, i.e.
$$
\frac{1}{2}(n+2n)(n+1)=\frac{1}{2}(3n^2+3n)
$$
dividing by $n^2$ gives
$$
\frac{1}{2}(3+3/n)
$$
As $n \to \infty$ the last term $3/n\to 0$ so the limit equals $3/2$.
