I am trying to solve problem 1.2 of Negele's book "Quantum many-particle systems":
Equation (1.123) is:
I managed to show the first inequality. I am having trouble with the second. From the second expression, we get \begin{equation} \sum_n\sum_m\int\frac{1}{\pi}d\rho d\theta\,\rho \frac{e^{-\rho^2}\rho^{n+m}e^{i\theta(n-m)}}{n!m!}\vert n\rangle\langle m\vert \end{equation}
Integrating over theta,
\begin{equation} \sum_n\sum_m\int\frac{1}{\pi}d\rho \, \frac{e^{-\rho^2}\rho^{n+m+1}\delta_{n,m}}{n!m!}\vert n\rangle\langle m\vert \end{equation}
That is \begin{equation} \sum_n\int\frac{1}{\pi}d\rho \, \frac{e^{-\rho^2}\rho^{2n+1}}{2n!}\vert n\rangle\langle n\vert \end{equation}
But now I don't know how to proceed. The integral should yield $\pi$, but I don't see how.
Your first step is not correct. You should have found \begin{align} \int\rho\frac{d\rho d\theta}{\pi}e^{-\rho^2}\sum_m\frac{(\rho e^{i\theta})^m}{\sqrt{m!}}\left|m\right>\sum_n\frac{(\rho e^{i\theta})^n}{\sqrt{n!}}\left|n\right|>&=\sum_{m,n}\int d\rho d\theta\frac{e^{-\rho^2}e^{i\theta(m-n)}\rho^{n+m+1}}{\sqrt{m!}\sqrt{n!}\pi}\left|m\right>\left<n\right| \\ &=2\sum_n\int d\rho\frac{e^{-\rho^2}\rho^{2n+1}}{\sqrt{n!}\sqrt{n!}}\left|n\right>\left<n\right| \\ &=2\sum_n\int d\rho\frac{e^{-\rho^2}\rho^{2n+1}}{n!}\left|n\right>\left<n\right|. \end{align} Here we have carelessly switched the sums and the integrals, carried out the integral over the angle to get $2\pi$ (do the computation carefully), and noted that $\sqrt{n!}\sqrt{n!}=n!$. Now the integral is a standard integral which you can integrate by parts or look up or ask Gandalf about, and its value is $n!/2$. This cancels $2/n!$, and the result follows.