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Assume that for each $n$, $X_n$ has a Poisson distribution with mean $\lambda_n = \sqrt{n\log{n}}$. We want to prove that

$$\lim_{n\rightarrow\infty} 1 - \sum_{i=0}^{\lfloor\sqrt{n}\rfloor} Pr(X_n=i) = 1$$

I would like to know if anybody has an idea how to prove this which is basically an asymptotic equality.

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We only need an asymptotic proof, so use Chebyshev's inequality despite being a loose bound:

$$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$$ which implies for the lower tail $$\Pr(\mu - X \geq k\sigma) \leq \frac{1}{k^2}.$$

Here $\mu=\lambda_n = \sqrt{n \log n}$ and with a Poisson distribution $\sigma=\sqrt{\lambda_n} = \sqrt[4]{n \log n}$.

We are interested in the case where $k\sigma = \sqrt{n \log n} - \sqrt{n}$, i.e. $k = \frac{\sqrt{n \log n} - \sqrt{n}}{\sqrt[4]{n \log n}} \ge \frac12\sqrt[4]{n \log n}$ for $n\ge e^4$ making $\Pr( X_n \leq \sqrt{n}) \leq \frac{4}{\sqrt{n \log n}}$, so $\Pr( X_n \leq \sqrt{n}) \to 0$ as $n \to \infty$ and so $$1 - \Pr( X_n \leq \sqrt{n}) \to 1 \text{ as } n \to \infty$$ which is equivalent to the expression in the original question.

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You can use Poisson lower tail bound (which corresponds to your summation) and show that it goes to 0 as n increases.

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