函数数列综合一题

nash

此题还不错

战巡

回复 1# nash


第一问太2了就不做了

\[\abs{a_k-a_{k+1}}=\abs{f(a_{k-1})-f(a_k)}\le L\abs{a_{k-1}-a_k}\le L^{k-1}\abs{a_1-a_2}\]
\[\sum_{k=1}^n\abs{a_k-a_{k+1}}\le \sum_{k=1}^nL^{k-1}\abs{a_1-a_2}\le \sum_{k=1}^{\infty}L^{k-1}\abs{a_1-a_2}=\frac{1}{1-L}\abs{a_1-a_2}\]

\[\abs{A_k-A_{k+1}}=\abs{\frac{(k+1)(a_1+a_2+...+a_k)-k(a_1+a_2+...+a_k+a_{k+1})}{k(k+1)}}=\abs{\frac{a_1-a_{k+1}+a_2-a_{k+1}+...+a_k-a_{k+1}}{k(k+1)}}\]
其中:
\[\abs{a_i-a_{k+1}}=\abs{\sum_{j=i}^k(a_j-a_{j+1})}\le \sum_{j=i}^k\abs{a_j-a_{j+1}}\le \sum_{j=i}^kL^{j-1}\abs{a_1-a_2}=\frac{L^i-L^{k+1}}{L(1-L)}\abs{a_1-a_2}\]
\[\abs{A_k-A_{k+1}}=\abs{\frac{1}{k(k+1)}\sum_{i=1}^k(a_i-a_{k+1})}\le \frac{1}{k(k+1)}\sum_{i=1}^k\abs{a_i-a_{k+1}}\]
\[\le \frac{1}{k(k+1)}\sum_{i=1}^k\frac{L^i-L^{k+1}}{L(1-L)}\abs{a_1-a_2}=\frac{1-(k+1)L^k+k L^{k+1}}{k(k+1)(1-L)^2}\abs{a_1-a_2}
=\frac{1}{(1-L)^2}(\frac{1-L^k}{k}-\frac{1-L^{k+1}}{k+1})\abs{a_1-a_2}\]
\[\sum_{k=1}^n\abs{A_k-A_{k+1}}\le \sum_{k=1}^{\infty}\abs{A_k-A_{k+1}}\le \sum_{k=1}^{\infty}[\frac{1}{(1-L)^2}(\frac{1-L^k}{k}-\frac{1-L^{k+1}}{k+1})]\abs{a_1-a_2}=\frac{1}{1-L}\abs{a_1-a_2}\]

nash

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