# $\sum_{t=1}^x e^{-\frac{1}{t}}$ 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?

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.