Bernoulli Trial Waiting Time converging to Poisson Process Arrival Time

Question

I am trying to model the following: if we have an i.i.d. sequence of random variables $(X_k^{(n)})_{k\in\mathbb{N}}$ where each r.v. follows a $Ber(\lambda/n)$ distribution, then I want to find the limiting distribution of $T_i^{(n)}/n$ as $n\to\infty$, where $T_i^{(n)}$ is the waiting time until the $i$-th success in the sequence.
The probability that $T_i^{(n)}=k$ for $k\geq i$ is given by $${{k-1}\choose{i-1}}(\frac{\lambda}{n})^i(1-\frac{\lambda}{n})^{k-i}$$ since we want $i-1$ successful trials in the preceding $k-1$ trials, and another success on the $k$-th trial. Therefore, $T_i^{(n)}\sim NB(i,\lambda/n)$, the negative binomial/Pascal distribution.
Given how the Poisson Process can be thought of as the limiting case of the Bernoulli process, it would make intuitive sense if $T_i^{(n)}/n\to T_i$ as $n\to\infty$, where $T_i$ is the waiting time for the Poisson Process. We know that $T_i\sim\Gamma(i,\lambda)$ for a P.P. with parameter $\lambda$.
Hence, I would like to prove this convergence. How would I go about doing this (provided the claim is true at all)? I am aware of the following result: If $X\sim NB(i,p)$, $Y\sim\Gamma(i,1/p)$, then $X\to Y$ as $p\to 0$. This would seem to be in contradiction with what I am trying to show. I am just not sure how to prove this, with the scaling term $1/n$ messing up the usual methods that I would try to use.