Exercise: Why $|\Theta|$ grows as expiry approaches

For an at-the-money Black-Scholes call, the gamma-driven component of theta is

\Theta_{\gamma} = -\frac{S\,\varphi(d_1)\,\sigma}{2\sqrt{\tau}}.

Tasks

At-the-money ( $S = K$ ) and for small $r$ , $d_1 \approx \tfrac{1}{2}\sigma\sqrt{\tau}$ . Show that $\varphi(d_1) \approx \varphi(0) = 1/\sqrt{2\pi}$ when $\sigma\sqrt{\tau}$ is small.
Using that approximation, conclude that $\Theta_{\gamma} \approx -\dfrac{S\sigma}{2\sqrt{2\pi\tau}}$ .
You own a $1$ -month ATM call and a $1$ -week ATM call on the same stock, both on $\$ 100 $with$ \sigma = 20%$. By what factor does the per-day gamma-driven theta of the 1-week call exceed that of the 1-month call?
Interpret this result in the language of gamma: why does theta grow like $1/\sqrt{\tau}$ near expiry?

Use the Black-Scholes ATM relations $\Gamma = \varphi(d_1)/(S\sigma\sqrt{\tau})$ and $\Theta_\gamma = -\tfrac12 \sigma^2 S^2 \Gamma$ , and compare.