Question

求解一道高数极限题，求过程

Answer 1

令A=∫(0→π/2)lncosxdx
B=∫(0→π/2)lnsinxdx
则B=∫(π/2→0)lnsin(π/2-t)d(π/2-t) (t=π/2-x)
=∫(0→π/2)lncosxdx
=A
2A=A+B=∫(0→π/2)ln(sinxcosx)dx
=∫(0→π/2)(ln(sin2x)-ln2)dx
=1/2*∫(0→π/2)lnsin(2x)d(2x)-πln2/2
=1/2*∫(0→π)lnsinxdx-πln2/2
=1/2*(∫(0→π/2)lnsinxdx+∫(π/2→π)lnsinxdx)-πln2/2
=1/2*(∫(0→π/2)lnsinxdx+∫(0→π/2)lnsintdt)-πln2/2 (t=π-x)
=∫(0→π/2)lnsinxdx-πln2/2
=A-πln2/2
所以A=-πln2/2

追问

求帮帮忙

追答

设∑(n=1→∞)(n+1)(u[n]-u[n+1])=l
任意正数a，存在正整数N0，使得当N>N0时∑(n=1→N)(n+1)(u[n]-u[n+1])∈(l-a,l+a)且Nu[N]∈(-a,a)
当N>N0时，
∑(n=1→N)(n+1)(u[n+1]-u[n])
=∑(n=1→N)(n+1)(-u[n]+u[n+1])
=-2u[1]+2u[2]-3u[2]+3u[3]-...-(N+1)u[N]+(N+1)u[N+1]
=-u[1]-∑(n=1→N)u[n]+(N+1)u[N+1]∈(l-a,l+a)
所以∑(n=1→N)u[n]∈(-l-u[1]-2a,-l-u[1]+2a)
即|∑(n=1→N)-(-l-u[1])|<2a
所以收敛

Answer 2

高中没毕业，老兄你难为我了