Answer
$3$
Work Step by Step
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}...=\Sigma_{p=0}^\infty\Sigma_{q=0}^\infty\dfrac{1}{2^p3^q}$
$\Sigma_{p=0}^\infty\Sigma_{q=0}^\infty\dfrac{1}{2^p3^q}=(\frac{1}{1-\frac{1}{2}})\times (\frac{1}{1-\frac{1}{3}})$
Hence, $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}...=3$
