<s>[INST] Limit of sequence as $n$ goes to infinity with $n$ in every term 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!
 [/INST] 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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