证明:若pk>o(k=1,2,……)(p是下标)且 lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)证明:若pk>o(k=1,2,……)(p是下标)且lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)则 lim{[p1an+p2a(n-1)+……+pna1]/p1+p2+……pn}=a.
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![证明:若pk>o(k=1,2,……)(p是下标)且 lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)证明:若pk>o(k=1,2,……)(p是下标)且lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)则 lim{[p1an+p2a(n-1)+……+pna1]/p1+p2+……pn}=a.](/uploads/image/z/10268721-9-1.jpg?t=%E8%AF%81%E6%98%8E%EF%BC%9A%E8%8B%A5pk%3Eo%28k%3D1%2C2%2C%E2%80%A6%E2%80%A6%29%EF%BC%88p%E6%98%AF%E4%B8%8B%E6%A0%87%EF%BC%89%E4%B8%94+lim%5Bpn%2Fp1%2Bp2%2B%E2%80%A6%E2%80%A6%2Bpn%5D%3D0%2Climan%3Da%28%E9%83%BD%E6%98%AFn%E2%86%92%E2%88%9D%29%E8%AF%81%E6%98%8E%EF%BC%9A%E8%8B%A5pk%3Eo%28k%3D1%2C2%2C%E2%80%A6%E2%80%A6%29%EF%BC%88p%E6%98%AF%E4%B8%8B%E6%A0%87%EF%BC%89%E4%B8%94lim%5Bpn%2Fp1%2Bp2%2B%E2%80%A6%E2%80%A6%2Bpn%5D%3D0%2Climan%3Da%28%E9%83%BD%E6%98%AFn%E2%86%92%E2%88%9D%29%E5%88%99++lim%7B%5Bp1an%2Bp2a%28n-1%29%2B%E2%80%A6%E2%80%A6%2Bpna1%5D%2Fp1%2Bp2%2B%E2%80%A6%E2%80%A6pn%7D%3Da.)
证明:若pk>o(k=1,2,……)(p是下标)且 lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)证明:若pk>o(k=1,2,……)(p是下标)且lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)则 lim{[p1an+p2a(n-1)+……+pna1]/p1+p2+……pn}=a.
证明:若pk>o(k=1,2,……)(p是下标)且 lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)
证明:若pk>o(k=1,2,……)(p是下标)且
lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)
则 lim{[p1an+p2a(n-1)+……+pna1]/p1+p2+……pn}=a.(极限是n→∝)
(注:1,2,……(n-1),n是下标)
证明:若pk>o(k=1,2,……)(p是下标)且 lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)证明:若pk>o(k=1,2,……)(p是下标)且lim[pn/p1+p2+……+pn]=0,liman=a(都是n→∝)则 lim{[p1an+p2a(n-1)+……+pna1]/p1+p2+……pn}=a.
把你要求极限的那个式子减去a,|p1an+p2a(n-1)+……+pna1]/(p1+p2+……pn)-a|<=
p1|(an-a)|+p2(|a(n-1)-a|)+……+pn(|a1-a|)]/(p1+p2+……pn)由前面给出的两个极限,可知任给ε,存在K,k大于K时,pn/p1+p2+……+pn<ε,|an-a|<ε.那么当n>2K+2时,把上面那个式子分成两部分,前半段由an的下标大于n/2-1的构成,后半部分由pn的下标大于n/2-1的构成,这样两部分之后将不小于原来的式子,对于第一部分,用|an-a|<ε来估计,对于第二部分,考虑到an是有界性,|an-a|