Question

Hi, can You tell me how to find the limit of [(1+2+3+...+n)/(2n^2 +3n)] where n tends to infinity? The answer is 1/4

Answer 1

mabye knowing a formula for the sum:\[1+2+\cdots +n\]would make the problem look a little easier...

Answer 2

\[1+2+3+\cdots +n=\frac{n(n+1)}{2}\]this is something very very useful. So now you problem is:\[\lim_{n\rightarrow \infty}\frac{n(n+1)}{2(2n^2+3n)}\]

Answer 3

Got it, Thanks!