Let $(\lambda_{mn})_{(m,\:n)\in\mathbb N^2}\subseteq[0,\infty)$ with $$\lambda_{mn}^2\le\mu_m\mu_n\;\;\;\text{for all }(m,n)\in\mathbb N^2\tag1$$ for some $(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$ such that $\sum_{n\in\mathbb N}\mu_n$ exists in $\mathbb R$. Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of a separable $\mathbb R$-Hilbert space $H$ and $$e_m\otimes e_n:=\langle\;\cdot\;,e_m\rangle_He_n\;\;\;\text{for }(m,n)\in\mathbb N^2\;.$$ Let $\mathfrak L_1(H)$ denote the space of nuclear operators on $H$.
How can we show that $$A:=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}e_m\otimes e_n$$ exists in $\mathfrak L_1(H)$?
I've tried the following: Let $\operatorname{HS}(H)$ denote the space of Hilbert-Schmidt operators on $H$. Note that $(e_m\otimes e_n)_{(m,\:n)\in\mathbb N^2}$ is an orthonormal basis of $\operatorname{HS}(H)$ and hence $A$ exists in $\operatorname{HS}(H)$ if and only if $\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2$ exists in $\mathbb R$. Now, \begin{equation}\begin{split}\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2&=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}^2\underbrace{\left\|e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2}_{=\:1}\\&\le\sum_{(m,\:n)\in\mathbb N^2}\mu_m\mu_n\\&=\left(\sum_{m\in\mathbb N}\mu_m\right)\sum_{n\in\mathbb N}\mu_n\end{split}\tag2\end{equation} and hence $A$ exists in $\operatorname{HS}(H)\supseteq\mathfrak L_1(H)$. However, I've no idea how I can show that $A$ even exists in $\mathfrak L_1(H)$.
