Sounds like a trivial consequence of Liouville's Approximation Theorem and the definition of Liouville's Numbers. See http://mathworld.wolfram.com/LiouvillesApproximationTheorem.html and http://mathworld.wolfram.com/LiouvilleNumber.html

Just write the numbers in base $ g$. For the first sum, the base-$ g$ expansion contains a $ 1$ at each $ g^{-n^2}$ place value, and zeros elsewhere. These ones are increasingly far apart, so the digits do not repeat nor terminate; it must therefore be irrational. The same method can be used for the second sum.

Let $f(n) : \mathbb{N} \rightarrow \mathbb{N}$ be any strictly increasing function with the property that $\limsup_n \left(f(n+1) - f(n)\right) = \infty$. Clearly this property holds for $f(n) = n^2$ and $f(n) = n!$. We claim that the sum $\displaystyle \sum_{i = 0}^{\infty} \frac{1}{g^{f(i)}}$ is irrational for any such function $f$. Assume by contradiction that there exist integers $p, q$ with $\displaystyle \frac{p}{q} = \displaystyle \sum_{i = 0}^{\infty} \frac{1}{g^{f(i)}}$ and let $p_n$ be such that $\displaystyle \frac{p_n}{g^{f(n)}} = \sum_{i = 0}^{n} \frac{1}{g^{f(i)}}$. Then choose $n$ large enough so that $f(n+1) - f(n) - 1 > \log_g{q}$. We then get the following contradiction: \begin{align*} \dfrac{1}{qg^{f(n)}} &\le \dfrac{p}{q} - \dfrac{p_n}{g^{f(n)}} \\ &= \displaystyle \sum_{i = n+1}^{\infty} \frac{1}{g^{f(i)}} \\ &\le \frac{1}{g^{f(n+1) - 1}} \displaystyle \sum_{i = 1}^{\infty} \frac{1}{g^i} \\ &\le \frac{1}{g^{f(n+1) - 1}}\displaystyle \sum_{i = 1}^{\infty} \frac{1}{2^i} \\ &= \frac{1}{g^{f(n+1) - f(n) - 1}g^{f(n)}} \\ &< \dfrac{1}{qg^{f(n)}} \end{align*}