Let \begin{align} f(d)&=\binom{d}{2}\left[\left( 1-\frac{2}{d}\right)^{d-2}\times \left(\frac{1}{d}\right)^2 \right]^2 \\ \\ &+\binom{d}{3}\left[\left( 1-\frac{3}{d}\right)^{d-3}\times \left(\frac{2}{d}\right)^3 \right]^2 \\ \\ &+\binom{d}{4}\left[\left( 1-\frac{4}{d}\right)^{d-4}\times \left(\frac{3}{d}\right)^4 \right]^2 \\ \\ &\vdots \\ &+ \binom{d}{d-1}\left[\left(\frac{1}{d}\right)\times \left(1-\frac{2}{d}\right)^{d-1} \right]^2 \end{align}
I am trying to check if there exists an upper bound $g(d)$ for $f(d)$ s.t. $g(d)\to 0$ as $d\to \infty$? Instead, if we can show that $f(d)\to 0$ as $d\to \infty$, that works for me.
The series looks complicated and am not sure where to start. Any ideas?
If, I am not mistaken, you are considering $$f(d)=\sum_{n=2}^{d-1} a_n\qquad \text{with} \qquad a_n=\binom{d}{n} \left(1-\frac{n}{d}\right)^{2 d-2 n} \left(\frac{n-1}{d}\right)^{2 n}$$ The $a_n$ are decreasing. For example $$\frac{a_{d-2}}{a_{d-1}}=8 \left(\frac{d-3}{d-2}\right)^{2 d}\,\, \frac{(d-2)^2 (d-1)}{(d-3)^4}$$ which, for large values of $d$ is $$\frac{a_{d-2}}{a_{d-1}}=\frac{8}{e^2 } \frac 1d\exp\Bigg[\frac{2}{d}+\frac{5}{6 d^2}+O\left(\frac{1}{d^3}\right) \Bigg]$$ while $$\frac{a_{d-3}}{a_{d-2}}=\frac{243}{16 e^2 }\frac 1d\exp\Bigg[\frac{3}{d}+\frac{10}{3 d^2} +O\left(\frac{1}{d^3}\right) \Bigg]$$ So, it seems that we cannot expect a very fast convergence.
From that, what it seems it that there is a limit for $d\times f(d)$. Looking at the "derivative", the maximum value is attained for $d=6$ and then $$d\times f(d) \leq \frac{203797}{7558272}\implies f(d) \leq \frac{203797}{7558272}\frac 1d$$
If you are only concerned by large values of $d$, use for the constant $0.0185$.