导数

二、基本初等函数求导公式欧拉常数(Euler-Mascheroniconstant)学过高等数学的人都知道，调和级数S=1+1/2+1/3+……是发散的，如下：由于ln(1+1/n)<1/n（n=1，2，3，…）于是调和级数的前n项部分和满足Sn=1+1/2+1/3+…+1/n>ln(1+1)+ln(1+1/2)+ln(1+1/3)+…+ln(1+1/n)=ln2+ln(3/2)+ln(4/3)+…+ln[(n+1)/n]=ln[2*3/2*4/3*…*(n+1)/n]=ln(n+1)由于limSn（n→∞）≥limln(n+1)（n→∞）=+∞所以Sn的极限不存在，调和级数发散。但极限S=lim[1+1/2+1/3+…+1/n-ln(n)]（n→∞）却存在，因为Sn=1+1/2+1/3+…+1/n-ln(n)>ln(1+1)+ln(1+1/2)+ln(1+1/3)+…+ln(1+1/n)-ln(n)=ln(n+1)-ln(n)=ln(1+1/n)由于limSn（n→∞）≥limln(1+1/n)（n→∞）=0因此Sn有下界而Sn-S(n+1)=1+1/2+1/3+…+1/n-ln(n)-[1+1/2+1/3+…+1/(n+1)-ln(n+1)]=ln(n+1)-ln(n)-1/(n+1)=ln(1+1/n)-1/(n+1)>ln(1+1/n)-1/n>0所以Sn单调递减。由单调有界数列极限定理，可知Sn必有极限，因此S=lim[1+1/2+1/3+…+1/n-ln(n)]（n→∞）存在。于是设这个数为γ，这个数就叫作欧拉常数，他的近似值约为0.57721566490153286060651209，目前还不知道它是有理数还是无理数。在微积分学中，欧拉常数γ有许多应用，如求某些数列的极限，某些收敛数项级数的和等。例如求lim[1/(n+1)+1/(n+2)+…+1/(n+n)]（n→∞），可以这样做：lim[1/(n+1)+1/(n+2)+…+1/(n+n)]（n→∞）=lim[1+1/2+1/3+…+1/(n+n)-ln(n+n)]（n→∞）-lim[1+1/2+1/3+…+1/n-ln(n)]（n→∞）+lim[ln(n+n)-ln(n)]（n→∞）=γ-γ+ln2=ln2                                                                                                                                                                                                                           设             均为正数，且                  则                                                                                                                                                                                                                                                 其中                              设                  则                             若                                                   则                                                                                                                                                                                                                                     求证(1+1/n)∧n<e<(1+1/n)∧（n+1）设数列a(n)=(1+1/n)^n，数列b(n)=(1+1/n)^(n+1)由于lim(x-->+oo)(1+1/x)^x=e得：lim(n-->+oo)a(n)=e；lim(n-->+oo)b(n)=lim(n-->+oo)a(n)*lim(n-->+oo)(1+1/n)=e.因此只需证数列a(n)单调递增且数列b(n)单调递减<1>证明数列a(n)单调递增：a(n)=(1+1/n)^n=(1+1/n)*(1+1/n)*…*(1+1/n)*1(n+1个因子相乘，运用不等式)<(((1+1/n)+(1+1/n)+…+(1+1/n)+1)/(n+1))^(n+1)=((n+2)/(n+1))^(n+1)=(1+1/(n+1))^(n+1)=a(n+1)即a(n)<a(n+1)，从而数列a(n)单调递增<2>证明数列b(n)单调递减：b(n)=(1+1/n)^(n+1)=1/(n/(n+1))^(n+1)=1/(1-1/(n+1))^(n+1)(令t=-(n+1)，换元)=(1+1/t)^t=a(t)由<1>得a(t)关于t单调递增，而t=-(n+1)关于n单调递减，由复合函数的单调性知，b(n)单调递减。由<1><2>,得证！