How far do I have to go for the tail of a binomial distribution with small $p$ to be $O(1/n)$? Let $n$ be a large integer, $p$ be a small number (say, $p=C/n$ for some constant $C \ll n$), and consider the tail of the binomial distribution $B(n,p)$, after $T$:
$$
\delta = \sum_{s=T}^{n} p^s (1-p)^{n-s} \binom{n}{s}
$$
I'd like to find a $T$ large enough that $\delta$ is on the order of $1/n$. Berry-Esseen doesn't work, as the error term in $O(1/\sqrt{n})$. I found a better approximation of these trucated binomial sum in this paper (Theorem 2), but it's pretty complicated, and I'm having trouble extracting any kind of asymptotic from it, let alone an analytical formula.
Surely, such a $T$ exists: if $T$ is close to $n$, $\delta$ is really tiny (inversely exponential in $n$); and if $T$ is close to $C$, then $\delta=O(1)$. I have no intuition, however, whether choosing e.g. $T=2C$ is enough to get to $\delta=O(1/n)$ or if requires a much larger $T$. Importantly, I'm not only interested in asymptotics; but I'd like to be able to give a reasonably small $T$ such that $\delta<1/n$.
 A: $\newcommand{\de}{\delta}$
We have 
\begin{equation}
 \de=\sum_{j=k}^n a_j, 
\end{equation}
where $k:=T$, 
\begin{equation}
 a_j:=a_{n,j;p}:=\binom nj p^j q^{n-j}, 
\end{equation}
$p=c/n$, $c:=C\in(0,\infty)$ (a constant), and $q:=1-p=1-c/n$. 
Suppose now that $n\to\infty$, $b\in(0,\infty)$ (a constant), and 
\begin{equation}
 k=k_n\sim b\frac{\ln n}{\ln\ln n}. 
\end{equation}
Then 
\begin{align*}
 a_k&=\frac{n(n-1)\cdots(n-(k-1))}{k!}\,p^k q^{n-k} \\ 
 &=\frac{(np)^k}{k!}\,e^{O(k^2/n)} (1-c/n)^{n-k} \\ 
 &=\frac{c^k}{k!}\,e^{O(1)}
 =\exp\{-(1+o(1))k\ln k\} \\ 
 &=\exp\{-(b+o(1))\ln n\}
 =n^{-b+o(1)}. 
\end{align*}
Also, for $j=k,k+1,\dots,n-1$
\begin{equation}
 \frac{a_{j+1}}{a_j}=\frac{n-j}{j+1}\,\frac pq\le\frac c{(j+1)(1-c/n)}=o(1). 
\end{equation}
So, by a geometric progression bound, 
\begin{equation}
 \de=\sum_{j=k}^n a_j\sim a_k=n^{-b+o(1)}. 
\end{equation}
We conclude that, to get $\de\asymp1/n$, we need exactly $b=1$, so that 
\begin{equation}
 k\sim\frac{\ln n}{\ln\ln n}. 
\end{equation}
