A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?

Question

The inequality $$ e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n, $$ where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.

Are there any more elementary proofs of this inequality?

What's the question exactly, that this is $e^1<=3$? Or for all n? — Alec Teal, Jan 01 '15 at 13:59
@AlecTeal The question is inequality for all $n$. (In hope that it may be an alternative way of obtaining the properties od $(e_n)$). — Przemysław Scherwentke, Jan 01 '15 at 14:01
I saw $e_n$ proven to be bounded by comparing it to $f_n=\left(1+\frac{1}{n}\right)^{n+1}$, where you get $e_n<e_{n+1}<f_{n+1}<f_n$, but this only helps you for $n\geq 9$. — Arthur, Jan 01 '15 at 14:03
@PrzemysławScherwentke post what you tried (so others can avoid the same mistake) — Alec Teal, Jan 01 '15 at 14:03
If you show that $e_n$ is increasing with limit $e\lt 2.75$ you have that this is true for $n\ge 4$ and just need to test the cases $1,2,3$ which are straightforward — Mark Bennet, Jan 01 '15 at 14:04
Would you please check over the two induction proofs given; I don't think either of them is correct. — user84413, Aug 30 '15 at 23:46

Huy · Accepted Answer · 2015-01-01T14:52:40.313

7

You can use induction.

For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,

$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \right) \left(1 - \frac{1}{n} \right) = 3 - \frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)}.$$

It is left to show

$$-\frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)} \leq -\frac{1}{n}$$

or equivalently

$$ -3n+3-n+1 \leq -n+1,$$

i.e. $n \geq 1$. Since $n-1 \in \mathbb{N}$, we have $n \geq 2$ by choice, so the inequality holds.

edited Jan 01 '15 at 14:52

answered Jan 01 '15 at 14:09

Huy

6,674

I have different inequality in the last line, but also true. +1. I am still waiting for another proofs, and then I'll choose the one for accepting. – Przemysław Scherwentke Jan 01 '15 at 14:39
3

By multiplying both sides of the second last inequality with $n(n-1)$, we get $-3(n-1) - n + 1 \leq -(n-1)$ which is equivalent to $-3n + 3 - n + 1 \leq -n + 1$, and by adding $3n-3$ to both sides, I ended up with $-n+1 \leq 2n-2$, if you're wondering. It seems simpler to just add $n-1$ to both sides, to retrieve $n \geq 1$ immediately. I changed the answer accordingly. – Huy Jan 01 '15 at 14:41
How did you get the factor of $1-\frac{1}{n}$ in the first line? – user84413 Aug 27 '15 at 18:20
I don't believe the inequality on the top line is correct for all n; for example, $(1+\frac{1}{3})^3=\frac{64}{27}>\frac{5}{3}=(3-\frac{1}{2})(1-\frac{1}{3})$. – user84413 Aug 30 '15 at 17:32
This answer is incorrect. The factor of $1+1/n$ magically changes to $1-1/n$, strengthening the proved inequality. – mjqxxxx Aug 20 '21 at 00:25
@mjqxxxx To me this seems more like a fatal flaw than strengthening the proved inequality. I don't see (at least at the moment) how this proof could be salvaged as if we proceed with $1+\frac{1}{n}$ - we get $3n-5 \leq 0$ which does not lead to the inductive step. – Figment Dec 17 '21 at 02:07

score 7 · Answer 2 · answered Jan 01 '15 at 15:03

7

Since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\leq\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\tag{1}$$ hence: $$ \left(1+\frac{1}{n}\right)^n \leq \exp\left(1-\frac{1}{2n+2}\right)\leq\frac{e}{1+\frac{1}{2n+2}}=\frac{2n+2}{2n+3}e \tag{2}$$ that is stronger than $ \left(1+\frac{1}{n}\right)^n \leq 3-\frac{1}{n}$ for any $n\geq 2$.

answered Jan 01 '15 at 15:03

Jack D'Aurizio

353,855

2

Very interesting method. +1. And now I have the problem, if I should accept this one or the most elementary one. :-) – Przemysław Scherwentke Jan 03 '15 at 05:11
I'm sorry to bother you about this, but how did you get the last inequality in (2)? – user84413 Aug 27 '15 at 00:56
@user84413: it is just a consequence of $e^x\geq 1+x$. – Jack D'Aurizio Aug 27 '15 at 12:08
Thanks a lot! Now I understand. – user84413 Aug 27 '15 at 18:12

score 4 · Answer 3 · answered Jan 01 '15 at 14:29

4

We can expand using Binomial Theorem:

$\displaystyle \left(1+\frac{1}{n}\right)^n = \sum\limits_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum\limits_{k=0}^{n} \frac{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{k!} \le \sum\limits_{k=0}^{n} \frac{1}{k!}$

Now, $\displaystyle \sum\limits_{k=0}^{n} \frac{1}{k!} \le 2 + \sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 3 - \frac{1}{n}$

which proves the required inequality.

answered Jan 01 '15 at 14:29

sciona

3,700

2

Yes, we can, but I am trying to find more elementary proof. – Przemysław Scherwentke Jan 01 '15 at 14:36
@PrzemysławScherwentke more elementary ? :) Then you must help me here, what kind of proof are you looking for and what all am I allowed to use so that you may call it elementary ? :-) – sciona Jan 01 '15 at 14:39
2

`The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.
Are there any more elementary proofs of this inequality?` I feel, that there must be a nonstandard proof. Well, one of them have already appeared.
– Przemysław Scherwentke Jan 01 '15 at 14:43
@PrzemysławScherwentke oops ! so this is the proof using Binomial Thereom that you mentioned in OP. Thanks ! I'll try think of a different way ! :) – sciona Jan 01 '15 at 14:45

A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?

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