Question

微积分求助

这个级数的和怎么求

Answer 1

把求和公式看做数列的通项公式，则该求和的结果等于n趋于无穷时数列的前n项和，即对数列的前n项和求极限，该数列可表示为An=3[1/(n+1)-1/(n+3)]，前n项和为Sn=5/2-3/(n+2)-3/(n+3)，当n趋于无穷时，Sn=5/2，即该求和项结果为5/2

Answer 2

6/[(n+1)(n+3)]
=3[ 1/(n+1)- 1/(n+3)]
∑(n=1->无穷) 6/[(n+1)(n+3)]
=3∑(n=1->无穷) [ 1/(n+1)- 1/(n+3)]
=3 { 1/2 +1/3 - lim(n->无穷)[ 1/(n+2)+ 1/(n+3)] }
=3 ( 1/2 +1/3 )
=5/2