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I need some help to prove this inequality:

$\ (1 + \frac{1}{a_n})^{a_n} \le (1 + \frac{1}{\lfloor{a_n}\rfloor})^{\lfloor{a_n}\rfloor + 1} $

where $\ n \in \mathbb{N} $ and$\ a_n \in \mathbb{R}$ with$\ a_n > 1 $ is any element of a sequence.

Stefano Balzarotti
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    I don't get it. There is no further conditions on $a_n$? I mean, e.g. with $a_3=8$ it doesn't work. – md5 Dec 27 '16 at 16:29
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    I don't get why you need the floor function if $n$ is a natural number. – scott Dec 27 '16 at 16:29
  • If $\lfloor n\rfloor =n$, why even include it in the inequality? – scott Dec 27 '16 at 16:31
  • sorry, I miss a_, in the second part of inequality – Stefano Balzarotti Dec 27 '16 at 16:32
  • Doesn't $LHS < e < RHS$ hold generally (at least for positive $a_n$) ? – Martin R Dec 27 '16 at 16:33
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    I believe you'll also need $a_n > 1$ for this to hold. – sTertooy Dec 27 '16 at 16:34
  • Yes, this is a part of an exercise to prove that sequence converges to e. The exercise ask to apply Squeeze theorem, to prove that sequence converges to e. I already proven the other part of inequality, but I am unable to prove this part... – Stefano Balzarotti Dec 27 '16 at 16:37
  • In the exercise there are no other constraints for $\ a_n $, but since this proof, is used for another proof with $\ a_n $ which tends to $\ +\infty $, we can require $\ a_n > 1 $ to make valid the inequality. – Stefano Balzarotti Dec 27 '16 at 17:06
  • You cant prove that the sequence converges to $e$ generally because the limit of this sequence is the definition of $e$. At most you want to prove that the sequence converge for some value between $2$ and $3$ what we call $e$. – Masacroso Dec 27 '16 at 17:15
  • In my book the the definition of e is the limit of $(1+\frac{1}{n})^n $.
    The exercise ask to prove that also $(1+\frac{1}{a_n})^{a_n} $ converges to e. The exercise ask to use this inequality: $\ (1 + \frac{1}{a_n + 1})^{\lfloor a_n \rfloor } \le (1 + \frac{1}{a_n})^{a_n} \le (1 + \frac{1}{\lfloor{a_n}\rfloor})^{\lfloor{a_n}\rfloor + 1} $ to give a proof.     – Stefano Balzarotti Dec 27 '16 at 17:29
  • Sorry the correct inequality is $\ (1 + \frac{1}{\lfloor{a_n}\rfloor + 1})^{\lfloor a_n \rfloor } \le (1 + \frac{1}{a_n})^{a_n} \le (1 + \frac{1}{\lfloor{a_n}\rfloor})^{\lfloor{a_n}\rfloor + 1} $

    I already proven with definition that both extremes of inequality converges to e. I was also able to prove that the first part of inequality is valid for any $\ a_n > 1 $, but I am unable to prove the second part of inequality.

     – Stefano Balzarotti Dec 27 '16 at 17:38

1 Answers1

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An Elementary Answer

Since $\lfloor x\rfloor+1\ge x$, we can use Bernoulli's Inequality: $$ \begin{align} \left(1+\frac1{\lfloor x\rfloor}\right)^{\lfloor x\rfloor+1} &=\left[\left(1+\frac1{\lfloor x\rfloor}\right)^{\large\frac{\lfloor x\rfloor+1}x}\right]^{\normalsize\,x}\\ &\ge\left(1+\frac{\lfloor x\rfloor+1}{\lfloor x\rfloor x}\right)^{\normalsize x}\\[9pt] &\ge\left(1+\frac1x\right)^x \end{align} $$ for $x\ge1$, so that $\frac1{\lfloor x\rfloor}$ exists.

Notes on Bernoulli's Inequality

At the end of this answer, Bernoulli's Inequality is proven for integer exponents using only induction.

At the end of this answer, the integer version is extended to the rational version using only induction.

Community
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robjohn
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  • Thanks, for sure this is a good solution, but the exercise ask to use the inequality with floor function. Furthermore the exercise in my book is before the chapter of derivative, so I think this is not the expected solution. So I give you an upvote, but I can't accept as an answer. – Stefano Balzarotti Dec 27 '16 at 18:26
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    For $a_n\ge1$, set $x=a_n$ and $y=\lfloor a_n\rfloor$, and you get the inequality you ask about – robjohn Dec 27 '16 at 18:30
  • Thanks for explanation, but I prefer to wait some time to see if someone can provide a solution without the use of derivatives. Otherwise I'll accept your answer. – Stefano Balzarotti Dec 27 '16 at 18:38
  • Thanks, I suppose the exercise requires the use of Bernoulli's Inequality. – Stefano Balzarotti Dec 27 '16 at 18:46
  • @StefanoBalzarotti: If you want an answer not using derivatives, it would be nice to specify that in the question. I have added an answer using only Bernoulli's Inequality. – robjohn Dec 27 '16 at 18:46