Does $\sum_{k =1}^n \Big(1 - \frac{1}{\lambda_n \mu_k} \Big)_+$ converge when $\mu_k \to 0$? Let $\{\mu_k\}$ be a positive, decreasing sequence, such that $\mu_k \to 0$ as $k \to \infty$.
For notion,  $x _+ = \max\{0, x\}$ denotes the positive part.
$\underline{\text{Series:}}$ $\quad$ We are going to define two series based on a sequence of numbers $\{\lambda_n\}$ and $\lambda$.
The series are
$$
S_n = \sum_{k =1}^n \frac{1}{\lambda_n} \Big(\lambda_n - \frac{1}{ \mu_k} \Big)_+, 
\quad \mbox{and} \quad 
S = \sum_{k =1}^\infty \frac{1}{\lambda} \Big(\lambda - \frac{1}{\mu_k} \Big)_+. 
$$
$\underline{\text{Choice of $\lambda_n, \lambda:$}}\quad$ The scalars $\lambda_n$ and $\lambda$ solve the following equations:
$$
\sum_{k = 1}^n \frac{1}{\mu_k}\Big(\lambda_n - \frac{1}{ \mu_k} \Big)_+ = 1, 
\quad \mbox{and} \quad 
\sum_{k = 1}^\infty  \frac{1}{\mu_k}\Big(\lambda - \frac{1}{ \mu_k} \Big)_+ = 1.
$$
Question:  Is it true that $S_n \to S$ as $n \to \infty$? The main difficulty is that the terms of the series $S_n$ depend on $\lambda_n$ rather than $\lambda$.

Following a suggestion of Steven Stadnicki.
$\mu_k$ are decreasing and so $1/\mu_k$ are increasing to $+\infty$.
Consequently, there exists $\kappa$ for which $\lambda < 1/\mu_\kappa$,
and thus $(\lambda - 1/\mu_k)_+ = 0$ for all $k \geq \kappa$.
Consequently,
$$
\sum_{k=1}^\infty \mu_k^{-1} (\lambda - \mu_k^{-1})_+ = 
\sum_{k=1}^{n} \mu_k^{-1}(\lambda - \mu_k^{-1})_+ 
$$
for all $n \geq \kappa - 1$.
This implies $\lambda = \lambda_n$ for all $n \geq \kappa - 1$.
Moreover, for $n \geq \kappa - 1$, it can be verified that
$$
S = S_n = \lambda^{-1} \sum_{k < \kappa} (\lambda - \mu_k^{-1}). 
$$
