# Limit value of a sequence [closed]

So, the quest is to determine the limit value of the following sequence: $$a_n= \frac{1}{n^2}\sum_{k=1}^{n} k$$ I don't even know how to start, could someone just give me a hint, please?

• It's an objective, "quest" is something from an epic, sounds very Middle-English like. Nov 18 '15 at 10:05
• "I don't even know how to start" This is definitely problematic, how on Earth are you in such a dysfunctional learning relationship that somebody is giving you exercises to solve that you "do not even know how to start"?
– Did
Nov 18 '15 at 10:19

You can prove by induction that $\sum_{k=1}^n k = \frac{n(n+1)}2$.
Start with $S_n=\sum_{k=1}^n=1+2+3+4+.....+n=\frac{n(n=1)}{2}$
Hint: You can easily prove that $\sum_{k=1}^n=\frac{n(n+1)}{2}$ which implies that $a_n=\frac{n+1}{2n}$. We can intuitively say that the limit is $\frac{1}{2}$ but you can do the same by writing it as $(\frac{1}{2}+\frac{1}{2n})$ and putting $n=\infty$. Now that we've found out the limit, we have to show that the limit actually is $\frac{1}{2}$. Take any $\epsilon>0$ then show that $\exists k$ such that $n>k \implies\left|(a_n-\frac{1}{2})\right|<\epsilon$. :)