Almost sure bounds on the maximum of Gamma distributions. For integers $n, m_n$, let $\Gamma_i(1, n)$ be i.i.d. random variables distributed like a Gamma distribution with a shape parameter $n$ and a scale parameter $1$. I am trying to find an almost sure upper bound on $\max_{1 \leq i \leq m_n} \Gamma_i(1, n)$ for $m_n \gg n$ (e.g. for $m_n = 2^n$) as $n \to \infty$.
Let
$$
S := \sum_{n = 0}^{\infty} \mathbb{P}(\max_{1 \leq i \leq m_n} \Gamma_i(1, n) > f(n)) =
\sum_{n = 0}^{\infty} (1 - \mathbb{P} \left( \Gamma_1(1, n) < f(n) \right)^{m_n})
\\
= \sum_n^{\infty} \left( 1 - \left( \frac{1}{(n-1)!} \int_0^{f(n)} s^{n-1} e^{-s} \mathrm{d} s\right)^{m_n} \right) \,.
$$
By Borel-Cantelli I need to find $f(n)$ such that $S < \infty$.
Now this integral isn't something I managed to bind well.
 A: It is more convenient to work with $\Gamma(1,n+1)$.

Claim. For $a>1$ $$P(\Gamma(1,n+1)>an)\le \frac{(an)^n}{e^{an}n!(1-1/a)}.$$

Proof
$$
P(\Gamma(1,n+1)>an) = \frac{1}{n!}\int_{an}^\infty x^n e^{-x} dx = \frac{1}{e^{an}n!}\int_{0}^\infty (an+z)^n e^{-z} dz  \\
= \frac{(an)^n}{e^{an}n!}\int_{0}^\infty \left(1+\frac{z}{an}\right)^n e^{-z} dz\le \frac{(an)^n}{e^{an}n!}\int_{0}^\infty e^{z/a} e^{-z} dz = \frac{(an)^n}{e^{an}n!(1-1/a)}.
$$

Now set $f(n) = a n$ with $a=a_n>1$ to be chosen later. Let $m_n = 2^n$. 
Then 
$$
\sum_{n = 1}^{\infty} {P}(\max_{1 \leq i \leq 2^n} \Gamma_i(1, n+1) > an) =
\sum_{n = 1}^{\infty} \Big(1 - \big(1-{P} ( \Gamma_1(1, n+1) >an )\big)^{2^n}\Big)
\\
\le  \sum_{n=1}^{\infty} \left( 1 - \left(1-\frac{(an)^{n} }{e^{an}n!(1-1/a)}\right)^{2^n}\right)\le \sum_{n=1}^{\infty} \frac{(2an)^{n} }{e^{an}n!(1-1/a)}\\
\le \sum_{n=1}^{\infty}\frac{(2ae)^n}{e^{an}(1-1/a)\sqrt{2\pi n}},
$$
where we have used the Bernoulli inequality and the fact that $n!\ge \sqrt{2\pi n} (n/e)^n$. 
Now set $a_n = b + k\frac{\log n}n$, where $b>1$ is such that $e^{b}/b = 2e$ (or $b - \log b = 1 + \log 2$); $k>k^* := \frac{b}{2(b-1)}$. Then the last series converges. 
Therefore, $f(n) = bn + k\log n$ with $k>k^*$ is the desired upper bound. It is in fact quite sharp (you can make it even sharper by taking $f(n) = bn + k^*\log n + r \log\log n$ with $r>$ something like $2k^*$). I have some reasons to believe that it is possible to prove divergence of the initial series for $k=k^*$.
Just make a comment if you have troubles proving the claim or with some steps in my argument.
