Do two matrices close in spectral norm have close traces? Let $A_p$ and $B_p$ be a sequence of symmetric positive-definite $p\times p$ matrices such that $$\|A_p - B_p\|_2 \le 1/p.$$ Suppose further that the spectra of $A_p$ and $B_p$, for all $p$, is contained in a compact interval in $(0,\infty)$. And finally suppose that for all $p$, $\operatorname{tr}(B_p) = p.$
Intuitively, it seems that it should follow from this that $$\|pA_p/\operatorname{tr}(A_p) - B_p\|_2 \le const./p.$$
Does this in fact hold?
 A: Yes, it does, and you don't need Weyl's inequality. Note that
$$
|\operatorname{tr}(A_p-B_p)|\le\sum_i|\lambda_i(A_p-B_p)|\le p\|A_p-B_p\|_2\le1.
$$
Therefore
$\operatorname{tr}(A_p)= p+\operatorname{tr}(A_p-B_p) \in [p-1,p+1]$ and
$$
\frac{-1}{p+1}=\frac{p}{p+1}-1\le\frac{p}{\operatorname{tr}(A_p)}-1\le\frac{p}{p-1}-1=\frac{1}{p-1}.
$$
In turn, $\left|\frac{p}{\operatorname{tr}(A_p)}-1\right|=O(\frac1p)$. Since we have assumed that the spectrum of $B_p$ is confined inside a compact interval, we also have $\|B_p\|_2=O(1)$. Thus
\begin{align*}
\left\|\frac{p}{\operatorname{tr}(A_p)}A_p-B_p\right\|_2
&\le\frac{p}{\operatorname{tr}(A_p)}\|A_p-B_p\|_2
+ \left|\frac{p}{\operatorname{tr}(A_p)}-1\right| \|B_p\|_2\\
&=O(1)O(\frac1p) + O(\frac1p)O(1) = O(\frac1p).
\end{align*}
A: To make a very slight generalisation: I'll prove this for Hermitian, positive-semidefinite matrices. 

Preliminaries:
From the inequality you give, and the fact that the spectra are contained in some compact interval $[\lambda_{min},\lambda_{max}]$, we can state:
$A_p$ and $B_p$ are positive-semidefinite Hermitian matrices, where $A_p$ has eigenvalues $\alpha_i$:
$$\lambda_{max} \geq \alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_p \geq \lambda_{min} \geq 0,$$
and $B_p$ has eigenvalues $\beta_i$:
$$\lambda_{max} \geq  \beta_1 \geq \beta_2 \geq \cdots \geq \beta_p\geq \lambda_{min} \geq 0,$$
then by the Weyl matrix inequalities: $\forall i = 1,\dots,p$:
$$|\alpha_i - \beta_i| \leq \|A_p-B_p\|_2 \leq 1/p.$$
Let us also note that the spectral norm $\|A_p\|_2$ of $A_p$ satisfies:
$$\|A_p\|_2 = \max_{i = 1,\,\dots\,,\,p}|\alpha_i| = \alpha_1 \leq \lambda_{max}.$$
Finally, note that for $p \geq 2$ we have that
$$\frac{1}{p-1} \leq \frac{2}{p}.$$

Claim: 

For all $p = 1, 2, \dots$ , we have that
  $$ \left\|\frac{p}{\operatorname{tr}(A_p)}A_p-B_p\right\|_2 \leq \frac{1+2\lambda_{max}}{p} $$

First of all, note that for $p=1$ the proof is quite trivial:
$$\frac{p}{\operatorname{tr}(A_p)}A_p = \frac{A_p}{A_p} = 1$$ 
and
$$B_p = \operatorname{tr}(B_p) = p = 1,$$
hence
$$ \left\|\frac{p}{\operatorname{tr}(A_p)}A_p-B_p\right\|_2 = \| 1 - 1 \|_2 = 0 \leq 1+2\lambda_{max}. $$
We will assume $p \geq 2$ from now on. Since $\operatorname{tr}(B_p) = p$ and $|\alpha_i - \beta_i| < 1/p$, we have that
$$|\operatorname{tr}(A_p)-\operatorname{tr}(B_p)| = \left|\sum_{i=1}^p \alpha_i - \sum_{i=1}^p\beta_i\right|=\left|\sum_{i=1}^p (\alpha_i - \beta_i)\right|  \leq \sum_{i=1}^p |\alpha_i - \beta_i| \leq \sum_{i=1}^p \frac{1}{p} = 1.$$
Define $\varepsilon = \operatorname{tr}(A_p) - \operatorname{tr}(B_p) \in [-1,1]$, such that $\operatorname{tr}(A_p) = \operatorname{tr}(B_p) + \varepsilon = p+\varepsilon > 0$.
Using our preliminaries, we get:
\begin{align} \left\|\frac{p}{\operatorname{tr}(A_p)}A_p-B_p\right\|_2 
&=\left\|\frac{p}{p+\varepsilon}A_p-B_p\right\|_2 \\
&=\left\|\frac{p + \varepsilon-\varepsilon}{p+\varepsilon}A_p-B_p\right\|_2 \\
&=\left\|A_p - B_p - \frac{\varepsilon}{p+\varepsilon}A_p\right\|_2 \\
&\leq \left\|A_p - B_p\right\|_2 + \left\|\frac{\varepsilon}{p+\varepsilon}A_p\right\|_2 \\
&\leq \frac{1}{p} + \frac{|\varepsilon|}{p+\varepsilon}\left\|A_p\right\|_2 \\
&\leq \frac{1}{p} + \frac{1}{p-1}\lambda_{max} \\
&\leq \frac{1}{p} + \frac{2}{p}\lambda_{max} \\
&= \frac{1+2\lambda_{max}}{p}.
\end{align}
