an+a(n+ 1)- 1 = 2√[ana(n+ 1)]，
That is [√an-√a(n- 1)]? = 1.
According to the title, {an} increases monotonically, a1>; 0,
∴√an-√a(n- 1)= 1
∴√an=n
That is an=n? .
For natural number n≥2, there is n? & gtn？ Tong -EN
So 1/n? & lt 1/[n(n- 1)]= 1/(n- 1)- 1/n
( 1/2)? +( 1/3)? +( 1/4)? +......+( 1/n)？ & lt 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+.......+ 1/(n- 1)- 1/n = 1- 1/n & lt； 1
So11+(1/2)? +( 1/3)? +( 1/4)? +......+( 1/n)？ & lt 1+ 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+.......+ 1/(n- 1)- 1/n = 2- 1/n & lt； 2
That is1/a1+1/a2+...+1/an
2.anbn=(n^3+n^2)/(n+3)^2
bn=(n+ 1)/(n+3)^2
b(n+ 1)-bn
=(-n^2-3n+2)/[(n+4)^2(n+3)^2]<； 0
It shows that bn is a decreasing sequence.
Namely b1>; B2> B3> B4> ............. & gt Billion.
Because b 1= 1/8.
b2 & lt 1/8
b3 & lt 1/8
So b 1+B2+B3+B4+ ... billion.
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