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1.32 Proof

(i) Let a=r-1a=r-1, so a>0a>0 and we can use the binomial theorem to expand

rn=(1+a)nr^{n}=(1+a)^{n}
=1+(n1)a+(n2)a2++an=1+{{n}\choose{1}}a+{{n}\choose{2}}a^{2}+\dots+a^{n}
=1+na+positive terms=1+na+{\hbox{positive terms}}
1+na;\geq 1+na;

and hence rnr^{n}\rightarrow\infty as nn\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 sns^{n}\rightarrow\infty as nn\rightarrow\infty, and rn=1/sn0r^{n}=1/s^{n}\rightarrow 0 as nn\rightarrow\infty.

Now when -1<r0-1<r\leq 0, we let t=-r;t=-r; so 0t<10\leq t<1 and rn=(-1)ntn0r^{n}=(-1)^{n}t^{n}\rightarrow 0 as nn\rightarrow\infty.