I am studying for an exam in introductory analysis and came across this exercise in a practice exam:

Determine the limit of $(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$.

I know that every single fraction in the sequence goes to $0$ so by the addition rule I thought maybe the entire sequence adds to $0$? My intuition tells me I'm wrong since there are infinite terms. I'm not sure where else to start.

Any hints would be appreciated. Thanks!

  • $\begingroup$ Every term goes to $0$ but the number of terms increases with $n$ $\endgroup$ – Ross Millikan Jun 16 '18 at 14:58
  • $\begingroup$ If every term goes to $0$ but the number of terms goes to $\infty$ then the limit may be some positive number. A simple example is this: $$ \underbrace{\frac 1 n + \cdots + \frac 1 n}_{3n \text{ terms}}. $$ The terms approach $0$ but the sum is always $3,$ so the limit as $n$ grows is $3.$ $\endgroup$ – Michael Hardy Jun 16 '18 at 14:59
  • $\begingroup$ You should not say "infinite terms" if you mean "infinitely many terms". An infinite term is a term that is infinite. If there were some of those, then there would be infinite terms, but there might not be infinitely many terms, for example if there were only six of them. $\endgroup$ – Michael Hardy Jun 16 '18 at 15:00
  • 1
    $\begingroup$ Take a look at this $(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2}) < (\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$. and the one on the left is simply $\frac{(n)(n+1)}{2n^2}$ which you can probably show goes to $1/2$ as $n \rightarrow \infty $ $\endgroup$ – Uday Khanna Jun 16 '18 at 15:01

We have,

$(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2})\le(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})\le(\frac{1}{n^2-n}+\frac{2}{n^2-n}+\frac{3}{n^2-n}+ \cdots + \frac{n}{n^2-n})\ \forall \ n \in N$

$\Rightarrow a_n= \frac{(n)(n+1)}{2(n^2)} \le (\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1}) \le \frac{(n)(n+1)}{2(n^2-n)} =b_n $

By L-Hopital's rule on the upper and lower bouding sequences we have $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\frac{1}{2}$ and therefore by Squeeze theorem we get

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


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