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Suppose that $(a_n)$ is a sequence of positive numbers only. It is to be proven that $\limsup \left(\frac { 1+a_{n+1 }}{ a_n}\right)^n\ge e$ and that this estimate cannot be improved.

I tried it as follows.

Suppose on the contrary, $\limsup \left(\frac { 1+a_{n+1 }}{ a_n}\right)^n\lt e$
For brevity, let $x_n=\left(\frac { 1+a_{n+1 }}{ a_n}\right)^n$ and also let $x^*=\limsup\left(\frac { 1+a_{n+1 }}{ a_n}\right)^n$

We choose an $s\gt 0$ such that $x^*\lt s\lt e$ and hence by $\limsup$ definition, it follows that there exists an $N\in \mathbb N$ such that $\begin{align}n\ge N\implies x_n\lt s&\implies \ln x_n\lt \ln s\\&\implies \ln (1+a_{n+1 })-\ln a_n\lt \frac{\ln s }n\\&\implies \ln (a_{n+1 })-\ln a_n\lt \frac{\ln s }n\\&\implies \ln(a_{N+k })-\ln a_N\lt \ln s \sum_{ i=N}^{ N+k-1}\frac 1i\lt \ln (N+k-1)+\gamma_{\text{Euler-Mascheroni}}+o(1)\\\end{align}$

How do I get contradiction from here?

My another question is: since the exercise specifically says that this estimate can't be improved, isn't it as good as saying that $x^*=e$? Thanks.

Koro
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  • The first part is answered here: https://math.stackexchange.com/q/417551/42969, and the second part is answered here: https://math.stackexchange.com/q/200679/42969. – Martin R Apr 26 '21 at 13:23
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    @MartinR: Thanks a lot. I have seen first one already and just now took a look at the one for second part shared by you, I don't understand it (the second part link you shared). Isn't there any other way? And also I want to know: is not able to improve the estimate is same as proving limsup is $e$? I think the answer is affirmative. – Koro Apr 26 '21 at 13:42

1 Answers1

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Suppose that $$ \limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\lt e\tag1 $$ $(1)$ means the for some $N$, $n\ge N$ implies $$ \left(\frac{1+a_{n+1}}{a_n}\right)^n\le e\tag2 $$ which is equivalent to $$ 1\le e^{1/n}a_n-a_{n+1}\tag3 $$ Multiply both sides of $(3)$ by $e^{-H_n}$, where the $H_n$ are the Harmonic Numbers. Then $$ e^{-H_n}\le e^{-H_{n-1}}a_n-e^{-H_n}a_{n+1}\tag4 $$ Summing $(4)$ yields a telescoping sum, resulting in $$ \sum_{n=N}^{M-1}e^{-H_n}\le e^{-H_{N-1}}a_N-e^{-H_{M-1}}a_M\tag5 $$ Since $e^{-H_{M-1}}a_M\ge0$, $(5)$ implies that $$ a_N\ge e^{H_{N-1}}\sum_{n=N}^{M-1}e^{-H_n}\tag6 $$ Since $\log(n+1)\le H_n\le\log(n)+1$, $(6)$ says $$ a_N\ge N\sum_{n=N}^{M-1}\frac1{en}\tag7 $$ and since the Harmonic Series diverges, $(7)$ says that $a_N$ cannot be a finite number.

This is a contradiction.

robjohn
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  • Sir, thanks a lot for answering my question. I'm going through it. Shouldn't there be Euler Mascheroni constant here $\ln n\le H_n\le \ln n+1$? I say so because $H_n -\ln n=\gamma +o(1)$ for big enough $n$'s. – Koro Apr 29 '21 at 10:33
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    While it is true that $\lim\limits_{n\to\infty}(H_n-\log(n))=\gamma$, it is also true that $\log(n)\le H_n\le\log(n)+1$. All that means is that $0\le\gamma\le1$. – robjohn Apr 29 '21 at 10:35
  • I never really though about value of $\gamma$. I unconsciously thought it to be between 2 and 3 like $e$. – Koro Apr 29 '21 at 10:37
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    $\gamma=0.57721566490153286060651209$ approximately – robjohn Apr 29 '21 at 10:39
  • Yeah. I figured after your first comment :) Thanks a lot :) – Koro Apr 29 '21 at 10:39
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    I show one method to compute $\gamma$ in this answer. The link to the 10000 digits of $\gamma$ is broken temporarily, however. – robjohn Apr 29 '21 at 10:43