# Annoying limit of suprema, involving arbitrary square-summable sequence

For reasons not really worth going into, I'm looking at the following limit ($n$ is always an integer): $$\lim_{\lambda\to0^+} \frac{2}{\lambda^2}\left[\max\left\lbrace \lambda d_0, \sup_{n \ge 1} \lambda\left(1 + \frac{1}{n}\right) d_n - \frac{1}{n} - \frac{1}{2n^2} \right\rbrace - \max\lbrace \lambda d_0, 0\rbrace\right],$$ where $(d_n)_{n=0}^\infty$ is a real square-summable sequence (i.e. $\sum_{n=0}^\infty d_n < \infty$). My question is,

Is this limit always $0$, regardless of $(d_n)$? If not, can we at least guarantee that the limit exists?

I've observed that $\sup_{n \ge 1} \lambda\left(1 + \frac{1}{n}\right) d_n - \frac{1}{n} - \frac{1}{2n^2} \ge 0$, as $0$ is always the limit of the sequence, hence the whole expression is non-negative. To show the limit always exists, the expression being monotone decreasing would definitely suffice, but even though the sequences I've tried all seem to satisfy this, I can't help but shake the feeling that it won't be true in general.

For this limit to be non-existent or non-zero, we require, for any fixed $\lambda > 0$, $\sup_{n \ge 1} \lambda\left(1 + \frac{1}{n}\right) d_n - \frac{1}{n} - \frac{1}{2n^2} > 0$. In particular, we need $$\lambda \left(1 + \frac{1}{n}\right)d_n - \frac{1}{n} - \frac{1}{2n^2} > 0 \iff d_n > \frac{1}{\lambda n}\left(1 - \frac{1}{2(n + 1)}\right)$$ for some $n$, depending on $\lambda$. As $\lambda$ grows smaller, the potential candidates for $n$ can only decrease. Could I use this and square summability to come to the conclusion?

I've tried looking at various examples (all which seem to tend monotonically to $0$), but I'm a bit stuck. Does anyone have any insight?