求和数列填空：1,1,1,2,3,5,9,（ ）

s(n)=1^2/2^(1-1)+2^2/2^(2-1)+...+n^2/2^(n-1),
s(n)/2=1^2/2^(2-1)+2^2/2^(3-1)+...+(n-1)^2/2^(n-1)+n^2/2^n,
s(n)/2=s(n)-s(n)/2=1^2/2^(1-1)+[2^2-1^2]/2^(2-1)+...+[n^2-(n-1)^2]/2^(n-1)-n^2/2^n
=1+(2*1+1)/2+(2*2+1)/2^2+...+[2*(n-1)+1]/2^(n-1)-n^2/2^n
=2[1/2+2/2^2+...+(n-1)/2^(n-1)]+[1+1/2+...+1/2^(n-1)]-n^2/2^n
=2t(n)+[(1-1/2^n)/(1-1/2)]-n^2/2^n
=2t(n)+2[1-1/2^n]-n^2/2^n
t(n)=1/2+2/2^2+...+(n-1)/2^(n-1),
t(n)/2=1/2^2+2/2^3+...+(n-2)/2^(n-1)+(n-1)/2^n,
t(n)/2=t(n)-t(n)/2=1/2+(2-1)/2^2+...+[(n-1)-(n-2)]/2^(n-1)-(n-1)/2^n
=1/2+1/2^2+...+1/2^(n-1)-(n-1)/2^n
=[1-1/2^n]/[1-1/2]-1-(n-1)/2^n
=2[1-1/2^n]-1-(n-1)/2^n,
s(n)=2*s(n)/2=2*[2t(n)+2[1-1/2^n]-n^2/2^n]
=4t(n)+4[1-1/2^n]-n^2/2^(n-1)
=8*t(n)/2+4[1-1/2^n]-n^2/2^(n-1)
=8{2[1-1/2^n]-1-(n-1)/2^n}+4[1-1/2^n]-n^2/2^(n-1)
=16[1-1/2^n]-8-8(n-1)/2^n+4[1-1/2^n]-n^2/2^(n-1)
=12-[10+4(n-1)+n^2]/2^(n-1)

求和【1,+∞】∑(n+1)/2ⁿ
S‹n›=2/2¹+3/2²+4/2³+5/2⁴+6/2⁵+.+n/2ⁿ⁻¹+(n+1)/2ⁿ.(1)
(1/2)S‹n›=2/2²+3/2³+4/2⁴+5/2⁵+6/2⁶+.+n/2ⁿ + (n+1)/2ⁿ⁺¹.(2)
(1)-(2)【错项相减】得：
(1/2)S‹n›=1+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶+.+1/2ⁿ-(n+1)/2ⁿ⁺¹
=1/2+1/2+1/2²+1/2³+1/2⁴+1/2⁵+1/2ⁿ-(n+1)/2ⁿ⁺¹
=1/2+(1-1/2ⁿ)-(n+1)/2ⁿ⁺¹
故S‹n›=1+2(1-1/2ⁿ)-(n+1)/2ⁿ
故S=n→+∞limS‹n›=n→+∞lim[1+2(1-1/2ⁿ)-(n+1)/2ⁿ]=n→+∞lim{1+2(1-1/2ⁿ)-[1/(2ⁿln2)]}=3.

当n很大时,有：1+1/2+1/3+1/4+1/5+1/6+...1/n = 0.57721566490153286060651209 + ln(n)//C++里面用log(n),pascal里面用ln(n)
0.57721566490153286060651209叫做欧拉常数
to GXQ:
假设;s(n)=1+1/2+1/3+1/4+..1/n
当 n很大时 sqrt(n+1)
= sqrt(n*(1+1/n))
= sqrt(n)*sqrt(1+1/2n)
≈ sqrt(n)*(1+ 1/(2n))
= sqrt(n)+ 1/(2*sqrt(n))
设 s(n)=sqrt(n),
因为：1/(n+1)

收敛数列可求和.
有些非收敛数列可相应的求出前N项和
数列求积有什么意义呢?