Proving Limit: lim n->inf (2n)/(2n+1) = 1

[Dell]

how do i prove this limit?

lim
n->inf $$\frac{0+2+4+...2n}{1+3+...(2n+1)}$$=1

i think that its because of the fact that 2n is so big compared to the rest, and compared to the 1 in (2n+1) that their affect on the lim is none, is this correct?
how do i write this mathematically>

 

[AUMathTutor]

You could always find a closed form expression for the sum of the first n even integers and the first n odd integers, and then substitute those.

One could also note that the bottom line is always (n+1) more than the top line... and then get the expression for the even integers, and use l'hospital's rule and keep the high-order terms.