Recall for a $Gamma(n, \lambda)$ \begin{align} f_{X}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{\alpha - 1}}{\Gamma(\alpha)} \end{align} We can rearrange the terms of the given expression to get a $Gamma(n/2, 1/2)$ inside the integral \begin{align} &= \lim_{n \to \infty} \bigg[ \frac{1}{2}\frac{1}{2^{n/2 - 1}\Gamma(n/2)} \int_{n + \sqrt{2n}}^{\infty} \exp \bigg( -\frac{1}{2}t \bigg) t^{n/2 - 1} dt \bigg]\newline &= \lim_{n \to \infty} \bigg[ \int_{n + \sqrt{2n}}^{\infty} \frac{\frac{1}{2}e^{-\frac{1}{2}t} \big(\frac{1}{2}t \big)^{\frac{n}{2} - 1}}{\Gamma(n/2)} \bigg]\newline &= \lim_{n \to \infty} \bigg[ P \bigg(Gamma \bigg(\frac{n}{2}, \frac{1}{2} \bigg) \geq n + \sqrt{2n}\bigg) \bigg] \end{align} But, it is known that a $\chi^{2}_{n}$ has the same distribution as a $Gamma(\frac{n}{2}, \frac{1}{2})$ (see section) \begin{align} &= \lim_{n \to \infty} \bigg[ P(\chi^{2}_{n} \geq n + \sqrt{2n}) \bigg] \end{align} and, the mean and variance of a $X^{2}_{n}$ are $n$ and $2n$ respectively \begin{align} &= \lim_{n \to \infty} \bigg[ P \bigg(\frac{\chi^{2}_{n} - n}{\sqrt{2n}} \geq 1 \bigg) \bigg]\newline \end{align} Since $n \to \infty$, we can use central limit theorem for a standard normal variable $Z$ \begin{align} &= P(Z \geq 1) = 1 - P(Z < 1) = 1 - \Phi(1) = 1 - 0.84134 = 0.15866 \end{align}