Limit of $\min \{a\in \mathbb{N} : \sum_{i=1}^{a}\frac1{i}\geq n\}$ equals $e$

Question

Given $a_n = \min \{a\in \mathbb{N} : \sum_{i=1}^{a}\frac1{i}\geq n\}$ $\forall n \in \mathbb{N}$, I want to prove that:

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=e$$

For that purpose, I am trying to prove that $\log a_{n+1}-\log a_n \to 1$.

From the definition of $a_n$, we have that $1+\frac12+\ldots+\frac1{a_n}\geq n$, and we have:

$$n\leq 1+\frac12+\ldots+\frac1{a_n}\leq \int_1^{a_n+1}\frac1{x}dx = \log(a_n+1)$$

In the same way, for $a_{n+1}$ we have $n+1\leq \log(a_{n+1}+1)$.

Combining both inequalities we have:

$$\log (a_{n+1}+1)-\log (a_n+1) \geq 1$$

However, I don't know how to get the other inequality. Can somebody help me? Many thanks in advance!

Answer 1

Let's note as (harmonic series) $$H_{a_n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{a_n}$$ From the definition of $a_n$ we have $$H_{a_n} -\frac{1}{a_n}< n\leq H_{a_n} \Rightarrow 0< H_{a_n} - n< \frac{1}{a_n} \tag{1}$$

Given this we also have: $$\ln{a_n}<H_{a_n} -\frac{1}{a_n} < n\leq H_{a_n}<\ln{a_n}+1\Rightarrow \\a_n<e^n<a_n\cdot e \Rightarrow\\ \frac{1}{e^{n}}<\frac{1}{a_n}<\frac{1}{e^{n-1}} \tag{2}$$

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From $(1)$ and $(2)$ $$0< H_{a_n} - n< \frac{1}{e^{n-1}} \tag{3}$$

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A s a result, combining $0< H_{a_n} - n< \frac{1}{e^{n-1}}$ and $0< H_{a_{n+1}} - (n+1)< \frac{1}{e^{n}}<\frac{1}{e^{n-1}}$ $$|(H_{a_{n+1}}-(n+1))-(H_{a_{n}}-n)|=|H_{a_{n+1}}-H_{a_{n}}-1|<\frac{1}{e^{n-1}}$$ or $$\lim\limits_{n\to\infty} (H_{a_{n+1}}-H_{a_{n}})=1 \tag{4}$$

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And finally $$(H_{a_{n+1}}-H_{a_{n}})-(\ln{a_{n+1}}-\ln{a_{n}})=\\ (H_{a_{n+1}}-\ln{a_{n+1}})-(H_{a_{n}}-\ln{a_{n}})\to \gamma -\gamma =0, n\to\infty$$ which, from $(4)$ and $|x-y|\geq {\bigg |}|x|-|y|{\bigg |}$, means $$\ln{a_{n+1}}-\ln{a_{n}}\to 1, n\to\infty$$

Answer 2

For $n \geq 1$, let $$H_n = \sum_{k=1}^n \frac{1}{k}$$

For all $x \in \mathbb{R}$, let $f(x) = \min \lbrace n \in \mathbb{N} : H_n \geq x \rbrace$, and $g(x)= \min \lbrace n \in \mathbb{N} : \ln(n)+\gamma \geq x \rbrace$.

Let $\varepsilon > 0$. By a very classical result, you have $$H_n = \ln(n)+ \gamma + o(1),$$

therefore there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $$|H_n - \ln(n)-\gamma| < \varepsilon$$

Fix a $x \geq H_N +1$. Then for all $n \in \mathbb{N}$, $$n \geq f(x) \Longrightarrow H_n \geq x \Longrightarrow (H_n \geq x \text{ and } n \geq N) \Longrightarrow \ln(n)+\gamma \geq x-\varepsilon \Longrightarrow n \geq g(x-\varepsilon)$$

So $f(x) \geq g(x-\varepsilon)$. Similarly, for all $n \in \mathbb{N}$, $$n \geq g(x) \Longrightarrow \ln(n)+\gamma \geq x\Longrightarrow H_n \geq x-\varepsilon \Longrightarrow n \geq f(x-\varepsilon)$$

So $g(x) \geq f(x-\varepsilon)$. So we basically proved that $$\forall \varepsilon > 0, \exists A \in \mathbb{R}, \forall x \geq A, g(x-\varepsilon) \leq f(x) \leq g(x+\varepsilon)$$

In particular, $$\forall \varepsilon > 0, \exists A \in \mathbb{R}, \forall x \geq A, \frac{g(x+1-\varepsilon)}{g(x+\varepsilon)} \leq \frac{f(x+1)}{f(x)} \leq \frac{g(x+1+\varepsilon)}{g(x-\varepsilon)}$$

This implies that there exists a sequence $(\varepsilon_k)$ that tends to $0$ such that for all $k \in \mathbb{N}$, one has $$\frac{g(k+1-\varepsilon_k)}{g(k+\varepsilon_k)} \leq \frac{f(k+1)}{f(k)} \leq \frac{g(k+1+\varepsilon_k)}{g(k-\varepsilon_k)}$$

Finally, by definition of $g$, $$\frac{g(k+1-\varepsilon_k)}{g(k+\varepsilon_k)} =\frac{\lfloor \exp(k+1-\varepsilon_k-\gamma)\rfloor}{\lfloor \exp(k+\varepsilon_k-\gamma)\rfloor} \sim \frac{ \exp(k+1-\varepsilon_k-\gamma)}{\exp(k+\varepsilon_k-\gamma)} \rightarrow e$$

and similarly $$\frac{g(k+1+\varepsilon_k)}{g(k-\varepsilon_k)} \rightarrow e$$

So finally, $$\lim_{k \rightarrow +\infty} \frac{f(k+1)}{f(k)} = e$$