Question

级数求和

求级数∑(n=0,无穷)(n+1)/n!的和

Answer 1

解：（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+···)—[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( )相同形式好合并成一项 得
(1-x)S(x) =(x-1)/x*(x+1/2*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|<1；
把x=-x带入上式就是要求的关键级数了，带入后，两边同时除以(1-x)得：
S(x)=1/(1-x)+ 1/x*ln(1-x) |x|<1
（2）∑(n=1→无穷）（-1）^n*n/((n+1)*2^(n+1
=2S(-1/2)
=4/3—2ln3/2
≈0.5224
解答这类题的关键是化陌生级数为熟悉的泰勒级数或者格林级数来解决

Answer 2