1. 级数: ∑(1/(n(n+1)))
2. n=[1:1:100];
3. 求和函数
function sumArr=sn(n)
anY=1 ./ (n .* (n .+ 1));
sumArr=[];
len=length(anY);
for i=1: len
sub=[];
for m=1:i
sub = [sub,[m]];
end;
subArr=anY(:,sub);
sumSubArr = sum(subArr);
sumArr=[sumArr,[sumSubArr]];
end;
end;
4. 求和: sumArr=sn(n);
5. 绘图: plot(n,sumArr);
6.结论: 由图可判定级数收敛,和为1。