Answer
$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\approx1.87821$
Work Step by Step
If we write out the sum, we get:
$\sum_{k=1}^6\frac{1}{2k-1}=\frac{1}{2(1)-1}+\frac{1}{2(2)-1}+\frac{1}{2(3)-1}+\frac{1}{2(4)-1}+\frac{1}{2(5)-1}+\frac{1}{2(6)-1}=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\approx1.87821$
