Show the improper integral $\int^{\infty}_0 \frac 1 y e^{-y} dy$ doesn't converge.
Using Wolfram Alpha: http://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605 the result is as stated (divergent)
I know I can prove that $$\int^{\infty}_0 \frac 1 y e^{-y} dy$$ doesn't converge if I can find a divergent function $f(y): 0 \le f(y) \le \frac 1 y e^{-y}$ for $y \rightarrow \infty$.
Can someone come up with such a function ? I was thinking $f(y) = \frac 1 y$ because I know the series corresponding to $\{\frac 1 n\}$ is divergent, but this function doesn't have the desired property.
For $y>1$, the value $\frac{1}{y}e^{-y}$ is smaller than $e^{-y}$, and since the integral $\int_1^\infty e^{-y}dy$ converges, the integral $$\int_1^\infty\frac{1}{y}e^{-y}dy$$ also converges. That should tell you that the problem with the integral may lie elsewhere.