Indeed. I was more giving my version of the “cheating method” in the post, in which top and bottom were attacked separately. It can certainly be made simpler, as you suggest.

]]>Why do it in two steps when you can do it in one? The quotient $(1+8/n+100/n^2)/(1-35/2n^2)^2$ tends to $1$, hence is less than $2$ for sufficiently large $n$.

]]>I like that approach too — in fact, it’s pretty similar to the way I decided to handle the series in part (iii).

]]>E.g., we have , and . Since , then is some such that it’s less than for all . Similarly, since , there’s some such that it’s greater than for all .

Therefore, for all , we have the summand less than .

(If we’d had on top, say, then we’d point out that is eventually greater than , and that is eventually less than , making the ratio eventually greater than .)

I like this because it doesn’t require thinking of any bounds at all, and it fits well with the comments about “the terms decrease roughly like “.

]]>*Thanks very much — corrected now.*