Show that: $$ \forall n \in\Bbb N: e < \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) $$

Till now i've only worked out a couple of proof sketches, and I don't have an idea how to proceed with them.

Please note that this question has already been asked here. The answer there uses derivatives and integrals which i'm not allowed to use.

First sketch

Consider the sequence: $$ x_n = \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) $$

One of the ways to show what's required is to show that: $$ x_{n+1} \le x_n $$ namely the sequence is monotonically decreasing. Now given $n\in\Bbb N$ we may calculate $x_1$: $$ x_1 = \left(1+{1\over 1}\right)^1\left(1+{1\over 2\cdot 1}\right) = 3 $$

Consider the limit: $$ \lim_{n\to\infty}x_n = \lim_{n\to\infty} \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) = e $$

Now based on the fact that the sequence tends to $e$ and it is monotonically decreasing and $x_1 = 3$, then it should follow that: $$ \forall n\in\Bbb N: x_n \ge e $$ Here comes the hard part, I could't prove that $x_n$ is monotonically decreasing. I've considered the fraction: $$ {x_{n+1}\over x_n} = \left(1 - {1\over (n+1)^2}\right)^n \cdot \frac{2n(2n+3)(n+2)}{(2n+2)(2n+1)(n+1)} $$

Not sure how to show it is less than $1$.

Second sketch

This sketch is based on the idea from my previous question. Namely it has been shown there that: $$ e \le \frac{n+2}{(n+1)(n+1)!} + \sum_{k=0}^n {1\over k!} $$

It looks like: $$ e \le \frac{n+2}{(n+1)(n+1)!} + \sum_{k=0}^n {1\over k!} \le \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) \tag1 $$

In such case if we prove $(1)$ we are done. I've given it several tries but it gets ugly very soon.

The question is:

Is it possible to utilize any of those sketches to finish the proof? If not what would be the way to prove the inequality using anything before the definition of a derivative/Taylor series/intergrals

Thank you!

  • $\begingroup$ why not use the Principle of Mathematical Induction? It's just 4 lines if you use that :P XD :) $\endgroup$ – Abhas Kumar Sinha Jan 25 at 15:27
  • $\begingroup$ @AbhasKumarSinha I would appreciate if you could share that idea $\endgroup$ – roman Jan 25 at 15:42
  • $\begingroup$ I mean your question can be solved with PMI but not those sketches $\endgroup$ – Abhas Kumar Sinha Jan 25 at 15:46

The first sketch works.

We need to prove that $$\frac{x_n}{x_{n+1}}>1$$ or $$\frac{\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{2n}\right)}{\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{2n+2}\right)}>1$$ or $$\frac{(n+1)^n(2n+1)(n+1)^{n+1}(2n+2)}{n^n(n+2)^{n+1}2n(2n+3)}>1$$ or $$\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}>\frac{2n+3}{2n+1}$$ or$$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}>\frac{2n+3}{2n+1},$$ which is true because $$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}=\left(1+\frac{1}{n^2+2n}\right)^{n+1}>1+\frac{n+1}{n^2+2n}+\frac{(n+1)n}{2(n^2+2n)^2}>\frac{2n+3}{2n+1}.$$

  • $\begingroup$ Thank you for the answer, could you please elaborate on where the first inequality comes from? As for the second part you've used binomial expansion, right? $\endgroup$ – roman Jan 25 at 12:52
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    $\begingroup$ @roman I added something. See now. Yes, I used the binomial expansion. $\endgroup$ – Michael Rozenberg Jan 25 at 13:11
  • $\begingroup$ That's a lot of details, thank you so much! And i've been able to follow with the last part by binomial expansion. $\endgroup$ – roman Jan 25 at 13:38
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    $\begingroup$ +1, but I found the last step a slog to verify - did I miss a shortcut? $\endgroup$ – Calum Gilhooley Jan 25 at 19:52
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    $\begingroup$ @CalumGilhooley it also took me some time, had to expand to a polynomial of $4$-th degree. $\endgroup$ – roman Jan 25 at 21:23

$$\frac{\log(1+x)}{x}+\log\left(1+\frac{x}{2}\right)=1+\frac{5x^2}{24}-\frac{5x^3}{24}+\frac{59 x^4}{320}-\ldots=1+\sum_{m\geq 2}a_m x^m $$ for any $x\in(0,1)$, where the sequence $\{a_m\}_{m\geq 2}$ can be proved to have alternating signs and the sequence $\{|a_m|\}_{m\geq 2}$ can be checked to be weakly decreasing by induction on $m$. It follows that the LHS is $>1$ for any $x\in(0,1)$: your inequality follows by exponentiating both sides (with $x$ being chosen as $\frac{1}{n}$) and by checking $n=1$ as a separate case.


Another short proof exists, if you use $~\displaystyle\left(1+\frac{1}{n}\right)^{\frac{1}{2}}<1+\frac{1}{2n}~$ and $~\displaystyle e<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$

which is proofed in the note of my question here .

  • $\begingroup$ Hm, nice approach. Thanks for sharing this. $\endgroup$ – roman Feb 1 at 12:56
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    $\begingroup$ @Roman : You are welcome. It was a pleasure. :) $\endgroup$ – user90369 Feb 1 at 13:15
  • $\begingroup$ I was thinking of mentioning this approach in a comment on Jack D'Aurizio's answer, because he gave a very nice proof of the monotonicity of $(1+\frac{1}{n})^{n+\frac{1}{2}}$ here. $\endgroup$ – Calum Gilhooley Feb 1 at 14:27
  • $\begingroup$ @CalumGilhooley : Even if monotony isn't absolutely necessary here, it's of course always good and interesting to see different methods of proof. And so it's good to mentioned it. :) $\endgroup$ – user90369 Feb 1 at 15:29

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