(i) Let a=r-1a=r-1, so a>0a>0 and we can use the binomial theorem to expand
and hence rn→∞r^{n}\rightarrow\infty as n→∞n\rightarrow\infty.
(iii) The odd terms are all 11; whereas the even terms are all -1-1.
(ii) First suppose that 0<r<10<r<1, so s=1/r>1s=1/r>1; then sn→∞s^{n}\rightarrow\infty as n→∞n\rightarrow\infty, and rn=1/sn→0r^{n}=1/s^{n}\rightarrow 0 as n→∞n\rightarrow\infty.
Now when -1<r≤0-1<r\leq 0, we let t=-r;t=-r; so 0≤t<10\leq t<1 and rn=(-1)ntn→0r^{n}=(-1)^{n}t^{n}\rightarrow 0 as n→∞n\rightarrow\infty.