B组第二题,利用柯西准则判别下列级数的收敛性的两个小题,不用函数项级数,交错级数等知识,只用放缩发
B组第二题,利用柯西准则判别下列级数的收敛性的两个小题,不用函数项级数,交错级数等知识,只用放缩发证明,求详细过程,谢谢...
B组第二题,利用柯西准则判别下列级数的收敛性的两个小题,不用函数项级数,交错级数等知识,只用放缩发证明,求详细过程,谢谢
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(1)令ak=sinkx/2^k,则|∑(k=n to n+p)ak|=
|sinnx/2^n+sin(n+1)x/2^(n+1)+...+sin(n+p)/2^(n+p)|<=1/2^n+1/2^(n+1)+...+1/2^(n+p)=(1-1/2^(p+1))/(2^n(1-1/2))<1/2^(n-1)
对任意的ε>0,存在N=[-lnε/ln2]+2,当n>N时,对任意的n、p属于正整数,有|∑(k=n to n+p)ak|<ε
由Cauchy准则知,∑(n=1 to +∞)an收敛
(2)设该级数为∑(n=1 to +∞)bn
(观察知bn具有规律:当n=3k+1、3k+2时bn=1/n,当n=3k+3时bn=-1/n(k=0,1,...))
则|∑(k=3n+1 to 6n)bk|=|1/(3n+1)+1/(3n+2)-1/(3n+3)+...+1/(6n-2+1/(6n-1)+1/(6n)|>1/(3n+1)+1/(3n+4)+...+1/(6n-2)>1/(3n+3)+1/(3n+6)+...+1/(6n)=1/3[1/(n+1)+1/(n+2)+...+1/(2n)|>1/3[n/(2n)]=1/6
取ε0=1/7,则对于任意的N属于正整数,总存在n0=3n+1>N、m0=6n>n0,有|∑(k=n0 to m0)bk|>ε0
由Cauchy准则知,∑(n=1 to +∞)bn发散
|sinnx/2^n+sin(n+1)x/2^(n+1)+...+sin(n+p)/2^(n+p)|<=1/2^n+1/2^(n+1)+...+1/2^(n+p)=(1-1/2^(p+1))/(2^n(1-1/2))<1/2^(n-1)
对任意的ε>0,存在N=[-lnε/ln2]+2,当n>N时,对任意的n、p属于正整数,有|∑(k=n to n+p)ak|<ε
由Cauchy准则知,∑(n=1 to +∞)an收敛
(2)设该级数为∑(n=1 to +∞)bn
(观察知bn具有规律:当n=3k+1、3k+2时bn=1/n,当n=3k+3时bn=-1/n(k=0,1,...))
则|∑(k=3n+1 to 6n)bk|=|1/(3n+1)+1/(3n+2)-1/(3n+3)+...+1/(6n-2+1/(6n-1)+1/(6n)|>1/(3n+1)+1/(3n+4)+...+1/(6n-2)>1/(3n+3)+1/(3n+6)+...+1/(6n)=1/3[1/(n+1)+1/(n+2)+...+1/(2n)|>1/3[n/(2n)]=1/6
取ε0=1/7,则对于任意的N属于正整数,总存在n0=3n+1>N、m0=6n>n0,有|∑(k=n0 to m0)bk|>ε0
由Cauchy准则知,∑(n=1 to +∞)bn发散
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