求极限lim n趋向于无穷(1/n)*n次方根下(n+1)(n+2)⋯(n+n)
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求极限lim n趋向于无穷(1/n)*n次方根下(n+1)(n+2)⋯(n+n)
记原式=P,
P=[(n+1)(n+2)(n+3).(n+n)/n^n]^(1/n)
={[(n+1)/n][(n+2)/n][(n+3)/n].[(n+n)/n]}^(1/n)
=[(1+1/n)(1+2/n)(1+3/n).(1+n/n)]^(1/n)
取自然对数,
lnP=(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)]
设f(x)=ln(1+x),
则P=[f(1/n)+f(2/n)+...+f(n/n)]/n,
当n→∞时,
应用分部积分法可求得
则当n→∞时,lnP=ln(4/e),即P=4/e.
P=[(n+1)(n+2)(n+3).(n+n)/n^n]^(1/n)
={[(n+1)/n][(n+2)/n][(n+3)/n].[(n+n)/n]}^(1/n)
=[(1+1/n)(1+2/n)(1+3/n).(1+n/n)]^(1/n)
取自然对数,
lnP=(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)]
设f(x)=ln(1+x),
则P=[f(1/n)+f(2/n)+...+f(n/n)]/n,
当n→∞时,
应用分部积分法可求得
则当n→∞时,lnP=ln(4/e),即P=4/e.
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