Actually the question has more details than what it says in the title. Sorry about that I may described the question wrongly.


Let $X\_1^n, X\_2^n,\dots$ be i.i.d. Bernoulli random variables with parameter $\lambda/n$, i.e. $X\_1^n \overset{d}{=}\operatorname{Be}(\lambda/n)$ with fixed $\lambda > 0$. Consider $$
T\_i^n := \inf\{k : X\_1^n + \cdots + X\_k^n = i\}.$$
And I want to show that $$
\frac{T\_i^n}{n}\xrightarrow[n\to\infty]{d}\text{Gamma}(i,\lambda).$$


This confuse me since we know that the sum of Bernoulli random variables asymptotically converges to Poisson distribution and I don't see any relationship between Poisson and Gamma distribution.


Can anyone help me out?



Does sum of i.i.d. Bernoulli random variables with parameter $\lambda/n$ asymptotically converge to Gamma distribution?
$\newcommand\la\lambda$ $\newcommand\nt{\lfloor nt\rfloor}$ 
For any natural $i,n,k$, let $S^n\_k:=X^n\_1+\dots+X^n\_k$, with $S^n\_0:=0$. Then for any real $t>0$
\begin{align}
P(T^n\_i/n>t)&=P(T^n\_i>\nt) \\
&=P(S^n\_{\nt}<i)\to P(S\_{\la t} <i) \\ 
&=\frac{\la^i}{\Gamma(i)}\int\_t^\infty u^{i-1} e^{-\la u}\,du
\end{align}
(as $n\to\infty$), where $S\_{\la t}\sim Poisson(\la t)$; 
the convergence holds by the [Poisson limit theorem](https://en.wikipedia.org/wiki/Poisson_limit_theorem); 
the last displayed equality can be obtained by integrating by parts $i-1$ times. 


Thus indeed, the distribution of $T^n\_i/n$ converges to the gamma distribution with the shape paratemer $i$ and the rate $\la$. 


