 Solution

The moment generating function is $M_X(t)=\mathbb{E}(e^{tX})$ wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions $M_n$ are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to $(\lambda -t)x$ givesFor the waiting time to one head, independence of successive tosses gives $\mathrm{Pr}(Y=j)=p(1-p{)}^{j-1}$ for $j\ge 1$. Summing the geometric series givesThe waiting times between consecutive heads are independent copies of $Y$, so their sum $N$ has moment generating function $M_Y(t{)}^k$. ConsequentlyFor fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow 0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the continuity theorem applies. The limiting distribution isequivalently a chi-squared distribution with $2k$ degrees of freedom.