Question

求级数∑(n=0→无穷）n*x^n/(n+1)的和函数,并计算∑(n=1→无穷）...

求级数∑(n=0→无穷）n*x^n/(n+1)的和函数,并计算∑(n=1→无穷）（-1）^n*n/((n+1)*2^(n+1))

Answer 1

（1）令S(x)=∑(n=0→无穷）n*x^n/(n+1)则S(x)
=x/2
+2/3*x^2
+3/4*x^3
+···
+n/(n+1)*x^n
+···
(1)两边同乘x：xS(x)=1/2*x^2+
2/3*x^3+3/4*x^4+···+n/(n+1)*x^(n+1)+
···
(2)上面(1)-(2)得：(1-x)S(x)=1/2*x+
1/6*x^2+
1/12*x^3+···+1/n(n+1)*x^n+···=(1-1/2)*x+
(1/2-1/3)*x^2+
···+[1/n-1/(n+1)]*x^n+···=(x+1/2*x^2+···+1/n*x^n+···)—汪宴[1/2x+1/3*x^2+···+1/(n+1)*x^n+···]把上式
后面[
]内乘以x,外面除以x配成前面(
)相困樱银同形式好合并成一项
得(1-x)S(x)
=(x-1)/x*(x+1/颂冲2*x^2+···+1/n*x^n+···)—
1所以只需要求出x+1/2*x^2+···+1/n*x^n+··即可.考虑我们熟悉的泰勒级数：ln(1+x)=x-1/2*x^2+···(-1)^(n-1)*1/n*x^n+···
|x|