How to estimate such type of series? Consider $p \in(0,1)$ and series $\displaystyle \sum_{k=0}^{2n-1}\binom{n^2}{k} p^{n^2 -k}(1-p)^{k}$.
What is a best way to estimate behaviour of this function such type of sums in term of $p$ and $n$? Actually I'm interested in $p = f(n)$, for which this series goes to $1$ with $n$ goes to infinity.
Maybe it's looks strange but I meet this problem in my probability task. Also, maybe there is some ideas for general case ($m \in \mathbb{N}$ instead of $2n-1$)?
 A: From a probabilistic point of view, we could view this sum as $\mathbb{P}(X_n\leq 2n-1)$ with $X_n\sim \text{Bin}(n^2,p)$. Now, as $n$ grows, its expectation $p n^2$ grows with the square of $n$ and the standard deviation $n\sqrt{p(1-p)}$ grows with the order $n$. This already gives us an indication that $X_n$ should grow as a stochastic sequence with the order $n^2$. In particular, we have by the central limit theorem
$$ \frac{X_n - pn^2}{n\sqrt{p(1-p)}} \stackrel{\mathcal{D}}{\longrightarrow} N(0,1) $$
Now, we can calculate our probability as $n$ tends to infinity
$$ \mathbb{P}(X_n \leq 2n-1) = \mathbb{P}\left(\frac{X_n - pn^2}{n\sqrt{p(1-p)}}\leq \frac{2n-1 - pn^2}{n\sqrt{p(1-p)}}\right) = \mathbb{P}\left(Z\leq \frac{2-1/n - p n}{\sqrt{p(1-p)}}\right) $$
As $n$ grows, the right hand within the probability tends to $-\infty$ and thus as expected, the overall probability will tend to $0$. To be precise, we can estimate this using upper-tail-inequalities provided $pn + 1/n > 2$.
$$ \mathbb{P}\left(Z\leq \frac{2-1/n - p n}{\sqrt{p(1-p)}}\right) \leq \frac{\exp\left(-\left(\frac{p n + 1/n - 2}{\sqrt{p(1-p)}}\right)^2/2\right)}{\frac{p n + 1/n - 2}{\sqrt{p(1-p)}}\cdot\sqrt{2}} $$
This method relies on how well the binomial distribution is estimated with the normal distribution. I am sure this goes perfectly fine. However, if you need a rigorous proof you have to think about how quick this converges and if we can upper bound this convergence. However, I'll just limit my answer to the main part of your question.
If you have any questions, feel free to ask!
EDIT: This method gives a good approximation of the series as $n$ grows. However, I am not quite sure if it is an underestimation or overestimation due to the convergence rate (I have not given this a thought, so it might be fairly easy to solve). However, if you are not satisfied, try to look into Chebyshev's or Markov's inequality which do not depend on normal distributions. Obviously they are a bit less tight but they don't depend on the Gaussian approximation and they will give you (with $100\%$ certainty) upper bounds.
A: This is not an answer but just the result from a CAS.
Considering
$$S_m=\sum_{k=0}^{m}\binom{n^2}{k}\, p^{n^2 -k}\,(1-p)^{k}$$ a CAS gives as a result
$$S_m=1-\binom{n^2}{m+1} (1-p)^{m+1}\, p^{n^2-m-1} \,\,
   _2F_1\left(1,-n^2+m+1;m+2;\frac{p-1}p\right)$$ where appears the Gaussian or ordinary hypergeometric function.
