급수의 수렴과 발산

Since $S>1,$ ${a}_{2}>0,$ and since $k<1,$ ${S}_{2}=1+{a}_{2}<1+(S-1)=S.$ If ${S}_{n}>S$ for some n, then there is a smallest n. For this n, $S>{S}_{n-1},$ so ${S}_{n}={S}_{n-1}+k(S-{S}_{n-1})$ $=kS+(1-k){S}_{n-1}<S,$ a contradiction. Thus ${S}_{n}<S$ and ${a}_{n+1}>0$ for all n, so ${S}_{n}$ is increasing and bounded by $S.$ Let ${S}_{*}=\lim \,{S}_{n}.$ If ${S}_{*}<S,$ then $\delta =k(S-{S}_{*})>0,$ but we can find n such that ${S}_{*}-{S}_{n}<\delta /2,$ which implies that ${S}_{n+1}={S}_{n}+k(S-{S}_{n})$ $>{S}_{*}+\delta /2,$ contradicting that ${S}_{n}$ is increasing to ${S}_{*}.$ Thus ${S}_{n}\to S.$