lim{[a^(1⼀n)+b^(1⼀n)]⼀2}^nn趋于无穷。能否利用均值不等式与单调性来证明？

在n>=1范围内，无法证明单调性
此题有多种解法，最简单的是利用重要极限和洛必达法则求解
原式=lim(n->∞) {{1+[a^(1/n)+b^(1/n)-2]/2}^{2/[a^(1/n)+b^(1/n)-2]}}^{[a^(1/n)+b^(1/n)-2]*n/2}
=lim(n->∞) e^{[a^(1/n)+b^(1/n)-2]*n/2}
令t=1/n，则t->0+
原式=lim(t->0+) e^[(a^t+b^t-2)/(2t)]
=lim(t->0+) e^[(lna*a^t+lnb*b^t)/2]
=e^[(lna+lnb)/2]

n趋于∞时ln{[a^(1/n)+b^(1/n)]/2}/(1/n)
趋于2/[a^(1/n)+b^(1/n)]*[a^(1/n)*lna+b^(1/n)*lnb]/2*(-1/n^2)/(-1/n^2)
趋于（lna+lnb）/2=ln√(ab)
原式=lime^{ln{[a^(1/n)+b^(1/n)]/2}/(1/n)},
=√(ab).

=lim((a^t+b^t)/2)^1/t
=e^limln(a^t/2+b^t/2)/t
=e^lim(a^t/2+b^t/2-1)/t
=e^limlna*a^t/2+lnb*b^t/2
=e^limlna/2+lnb/2
=√ab

（1）当n为偶数时，令n=2k，则k=n/2
Sn=1²-2²+3²-4²+……+(2k-1)²-(2k)²
=(1-2)(1+2)+(3-4)(3+4)+……+(2k-1-2k)(2k-1+2k)
=-1-2-3-4-……-(2k-1)-2k
=-(2k+1)*2k/2
=-k(2k+1)
=-n(n+1)/2
（2）当n为奇数时，令n=2k-1，则k=(n+1)/2
Sn=1²-2²+3²-4²+……+(2k-3)²-(2k-2)²+(2k-1)²
=(1-2)(1+2)+(3-4)(3+4)+……+(2k-3-2k+2)(2k-3+2k-2)+(2k-1)²
=-1-2-3-4-……-(2k-3)-(2k-2)+(2k-1)²
=-(2k-1)*(2k-2)/2+(2k-1)²
=k(2k-1)
=n(n+1)/2
综上所述，
Sn=(-1)^(n+1)*n(n+1)/2