Compute $\lim_{n\to \infty} \left({1 \over n^2} + {2 \over n^2} + \cdots + {n - 1 \over n^2}\right)$

Question

$$\lim_{n\to \infty} \left({1 \over n^2} + {2 \over n^2} + \cdots + {n - 1 \over n^2}\right)$$

I tried solving this by finding $d$ which is $a_2 - a_1 = d$, but I don't know how to continue with it because it goes to inifnity and $S_n$ i beilieve works only for an infinite set.

I also tried with sandwich, i mean $b_c \le a_n \le c_n$ but when I checked $b_n$ and $c_n$ limits they were not equal to one another so I couldn't find $a_n$ limit.

Would love to get some help, and sorry for my english, I study this on another language.

Hint: $\dfrac{1}{n^2}+\dfrac{2}{n^2}+\cdots+\dfrac{n-1}{n^2}=\dfrac{1+2+\cdots+(n-1)}{n^2}$. — Corellian, Nov 12 '15 at 20:01
Okay. How can you rewrite $1+2+\cdots+(n-1)$? What do you know about the limit of a rational function at infinity? — Corellian, Nov 12 '15 at 20:02
@Dr.MV You're correct. My apologies. I'll delete and restate my comment. — Corellian, Nov 12 '15 at 20:26
@Dr. MV Of course, thanks for pointing it out! So, we have $\dfrac{1}{n^2}+\dfrac{2}{n^2}+\cdots+\dfrac{n-1}{n^2}=\dfrac{n(n-1)}{2n^2}=(n^2-n)/(2n^2).$ From here, you can break up the quotient and simplify. What do you get? Or, you can use the fact that $\displaystyle\lim_{x\to\infty}\dfrac{a_k x^k+\cdots+a_0}{b_k x^k+\cdots+b_0}=a_k / b_k$. — Corellian, Nov 12 '15 at 20:32
@Brody You're welcome! My pleasure. ;-)) And +1 for your comment. — Mark Viola, Nov 12 '15 at 20:38
See also: Evaluating the limit $\lim_{n\to\infty} \left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right)$ — Martin Sleziak, May 28 '18 at 10:48

score 5 · Accepted Answer · answered Nov 12 '15 at 20:01

5

HINT:

$$\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$$

answered Nov 12 '15 at 20:01

Mark Viola

179,405

Still can't find how to get to $1 \over 2$ from here. – Ilan Aizelman WS Nov 12 '15 at 20:06
1

Can you find the limit $\lim_{n\to \infty}\frac{n(n-1)}{2n^2}=\lim_{n\to \infty}\frac12 \left(1-\frac1n\right)$? – Mark Viola Nov 12 '15 at 20:07
Well yes. but it is the limit in the left side. how can you find in the right side now? – Ilan Aizelman WS Nov 12 '15 at 20:09
The limits are equal. As $n$ grows to $\infty$, $1/n$ decreases to $0$. The limit is, therefore, $1/2$. – Mark Viola Nov 12 '15 at 20:16

score 5 · Answer 2 · answered Nov 12 '15 at 20:03

5

Alternatively, you may use a Riemann sum: $$ \dfrac{1}{n^2}+\dfrac{2}{n^2}+\cdots+\dfrac{n-1}{n^2}=\frac1{n}\sum_{k=0}^{n-1}\frac{k}{n} \to \int_0^1x\:dx=\color{red}{\frac12}. $$

answered Nov 12 '15 at 20:03

Olivier Oloa

120,989

score 2 · Answer 3 · answered Nov 12 '15 at 20:16

2

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

answered Nov 12 '15 at 20:16

Prabath Hewasundara

1,644

same answer :-) I just saw your answer..I was busy with typing this I am new to here. it takes some time to type.. – Prabath Hewasundara Nov 12 '15 at 20:22

Compute $\lim_{n\to \infty} \left({1 \over n^2} + {2 \over n^2} + \cdots + {n - 1 \over n^2}\right)$

3 Answers3

Linked

Related