Question

求数项级数的和∑ ((-1)^n )*(n+1)/(2n+1)!

n=0到正无穷，，，求解啊~~~~

Answer 1

∑ ((-1)^n )*(n+1)/(2n+1)!
=(1/2)∑ ((-1)^n )*(2n+2)/(2n+1)!
=(1/2)∑ ((-1)^n )*(2n+1)/(2n+1)!+(1/2)∑ ((-1)^n )/(2n+1)!
=(1/2)∑ ((-1)^n )/(2n)!+(1/2)∑ ((-1)^n )/(2n+1)!
=(cos1+sin1)/2

Answer 2

设该级数的和为S，因
2S = 2∑[n=0~inf.)[(-1)^n]*(n+1)/(2n+1)!
= ∑[n=0~inf.)[(-1)^n]*/(2n)! + ∑[n=0~inf.)[(-1)^n]*/(2n+1)!，
考虑
sinx = ∑[n=0~inf.)[(-1)^n][x^(2n+1)]*/(2n+1)!，
是不是可以求和了？……