$\sum_{t=1}^x e^{-\frac{1}{t}} $ approximating $\log_e(\pi(e^x))\sim x$ Related to a previous question: Is $\ln(\pi(e^x)) \sim x?$
$\sum\limits_{t=1}^x e^{-\frac{1}{t}} $ approximates a modified prime counting function $\ln(\pi(e^x))\sim x$.
This is similar I guess to $\frac{x}{\ln{x}}$ approximating $\pi(x)$.

But doesn't $\sum\limits_{t=1}^x e^{-\frac{1}{t}} $ converge to $\ln(\pi(e^x))?$

Is it really that good of an approximation?
 A: Indeed
$$\sum\limits_{t=1}^x e^{-\frac{1}{t}} \sim x \tag{1}$$
and as a result (the link you provided)
$$\sum\limits_{t=1}^x e^{-\frac{1}{t}} \sim \ln{\pi(e^x)}$$
But let's prove $(1)$ ...

First of all
$$\sum\limits_{t=1}^x e^{-\frac{1}{t}}=\sum\limits_{t=1}^{\left \lfloor x \right \rfloor} e^{-\frac{1}{t}} \tag{2}$$
and $$1-x\leq e^{-x}\leq 1,\forall x\geq0$$
As a result
$$\left \lfloor x \right \rfloor-\sum\limits_{t=1}^{\left \lfloor x \right \rfloor}\frac{1}{t}\leq \sum\limits_{t=1}^{\left \lfloor x \right \rfloor} e^{-\frac{1}{t}} \leq \left \lfloor x \right \rfloor$$
However
$$\left \lfloor x \right \rfloor-(\ln{\left \lfloor x \right \rfloor}+1)\leq\left \lfloor x \right \rfloor-\sum\limits_{t=1}^{\left \lfloor x \right \rfloor}\frac{1}{t}\leq \sum\limits_{t=1}^{\left \lfloor x \right \rfloor} e^{-\frac{1}{t}} \leq \left \lfloor x \right \rfloor \Rightarrow\\
1-\frac{\ln{\left \lfloor x \right \rfloor}+1}{\left \lfloor x \right \rfloor}\leq 
\frac{\sum\limits_{t=1}^{\left \lfloor x \right \rfloor} e^{-\frac{1}{t}}}{\left \lfloor x \right \rfloor} 
\leq 1$$
and taking the limit
$$\lim\limits_{x\to\infty} \frac{\sum\limits_{t=1}^{\left \lfloor x \right \rfloor} e^{-\frac{1}{t}}}{\left \lfloor x \right \rfloor}=1 \tag{3}$$
because $x\to\infty \Rightarrow \left \lfloor x \right \rfloor \to\infty$ (not too difficult to show).
As a result, combining $(2)$ and $(3)$
$$\sum\limits_{t=1}^x e^{-\frac{1}{t}}\sim \left \lfloor x \right \rfloor$$
But
$$\left \lfloor x \right \rfloor \sim x$$
and $(1)$ follows.
