Watch Your steps!

Firstly compute the closed form for $I_n(a)$ as follows. We use the fact that the integral looks similar to the Gamma Function and thus we want to manipulate it into it. $\begin{aligned} I_n(a)&=\int_0^\infty t^n a^{-t}\ \mathrm dt\\ (\text{Let}\ w=t \log a)\qquad &=\frac{1}{(\log a)^{n+1}}\int_0^\infty w^n e^{-t}\ \mathrm dw\\ &=\frac{\Gamma(n+1)}{(\log a)^{n+1}} \end{aligned}$

Therefore we can evaluate the sum

$\sum_{n=0}^\infty\frac1{I_n(\pi)}=\log\pi\ \sum_{n=0}^\infty\frac{(\log \pi)^n}{\Gamma(n+1)}=\log \pi\ \sum_{n=0}^\infty\frac{(\log \pi)^n}{n!}\\ =\pi \log \pi \approx\ \boxed{3.596}$ The last one being the Taylor series of $\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$