Question

已知各項均不相等的正項數(shù)列{a_n}，{b_n}的前n項和分別為S_n，T_n．
（1）若{a_n}，{b_n}為等差數(shù)列，求證：．
（2）將（1）中的數(shù)列{a_n}，{b_n}均換作等比數(shù)列，請給出使成立的條件．

Answer 1

解：（1）證明：設{a_n}，{b_n}的公差分別為d₁，d₂（d₁，d₂均不為0），則{^{lim}_{x→∞}}a_{n}/b_{n}={^{lim}_{x→∞}}a_{1}+(n-1)d_{1}/b_{1+(n-1)d_{2}}=d_{1}/d_{2}，…（4分）{^{lim}_{x→∞}}S_{n}/T_{n}{^{lim}_{x→∞}}na_{1}+{n(n-1)/2d_{1}}{nb_{1}+n(n-1)/2d_{2}}=d_{1}/d_{2}，所以{^{lim}_{x→∞}}a_{n}/b_{n}={^{lim}_{x→∞}}S_{n}/T_{n}．…（8分）（2）設{a_n}，{b_n}的公比分別為q₁，q₂（q₁，q₂均為不等于1的正數(shù)），則{^{lim}_{n→∞}}a_{n}/b_{n}={^{lim}_{n→∞}}a_{1}q_{1}^{n-1}/b_{1q_{2}^{n-1}}=a_{1}/b_{1}{^{lim}_{n→∞}}(q_{1}/q_{2})^{n-1}=\begin{cases} {a_{1}/b_{1}(q_{1}=q_{2})} \\ {0(q_{1}＜q_{2}).} \end{cases}…（11分）{^{lim}_{n→∞}}S_{n}/T_{n}=a_{1}(1-q_{2})/b_{1(1-q_{1})}{^{lim}_{n→∞}}1-q_{1}^{n}/1-q_{2^{n}}=\begin{cases} {a_{1}/b_{1}(q_{1}=q_{2})} \\ {a_{1}(1-q_{2})/b_{1(1-q_{1})}(0＜q_{1}＜1，0＜q_{2}＜1)} \\ {0(0＜q_{1}＜q_{2}，q_{2}＞1).} \end{cases}…（14分）所以使{^{lim}_{x→∞}}a_{n}/b_{n}={^{lim}_{x→∞}}S_{n}/T_{n}成立的條件是0＜q₁＜q₂，q₂＞1或q₁=q₂．…（16分）