设数列{an}满足a1=2,an+1=an+1/an,(n∈N).令bn=an/根号下n,判断bn与bn+1的大小a1=2a(n+1)=an+(1/an)a(n+1) > anb(n+1)-bn = a(n+1)/ √(n+1) - an/√n> an/ √(n+1) - an/√n<0b(n+1) < bn

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设数列{an}满足a1=2,an+1=an+1/an,(n∈N).令bn=an/根号下n,判断bn与bn+1的大小a1=2a(n+1)=an+(1/an)a(n+1) > anb(n+1)-bn = a(n+1)/ √(n+1) - an/√n> an/ √(n+1) - an/√n<0b(n+1) < bn

设数列{an}满足a1=2,an+1=an+1/an,(n∈N).令bn=an/根号下n,判断bn与bn+1的大小a1=2a(n+1)=an+(1/an)a(n+1) > anb(n+1)-bn = a(n+1)/ √(n+1) - an/√n> an/ √(n+1) - an/√n<0b(n+1) < bn
设数列{an}满足a1=2,an+1=an+1/an,(n∈N).令bn=an/根号下n,判断bn与bn+1的大小
a1=2
a(n+1)=an+(1/an)
a(n+1) > an
b(n+1)-bn = a(n+1)/ √(n+1) - an/√n
> an/ √(n+1) - an/√n
<0
b(n+1) < bn

设数列{an}满足a1=2,an+1=an+1/an,(n∈N).令bn=an/根号下n,判断bn与bn+1的大小a1=2a(n+1)=an+(1/an)a(n+1) > anb(n+1)-bn = a(n+1)/ √(n+1) - an/√n> an/ √(n+1) - an/√n<0b(n+1) < bn
a1=2,an+1=an+1/an,(n∈N).
bn=an/√n,
∴b-bn=(an+1/an)/√(n+1)-an/√n2k+1/2,那么
a^2=(ak+1/ak)^2>ak^2+2>2k+1/2+2=2(k+1)+1/2,
即n=k+1时①也成立,
∴对任意n∈N+,①成立
∴b