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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?

rtybase
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John Zimmerman
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1 Answers1

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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.

rtybase
  • 16,907