How can I prove $\left(1+{1\over n}\right)^{n}\le\left(1+{1\over 1+n}\right)^{n+1}$ via induction?

Question

I am trying to prove this: $$\left(1+{1\over n}\right)^{n}\le\left(1+{1\over 1+n}\right)^{n+1}$$

I tried to prove by changing the main problem for a more general one , namely: $$a^{n}\le b^{n+1}$$ for $a$ and $b$ integers $>1$ and $a \le b$. Then by induction I tried to prove that (an induction on the three variables $a, b$ and $n$). For base case I chose $n=1 , b=2$ and $ a=3. $ After that I used bernouilli to get from $$(a+1)^{n+1}\le(b+1)^{n+2}$$ to $$1+a(n+1)\le 1+b(n+2)$$ then removing $1$ : $$a(n+1)\le b(n+2)$$ and after development $$an+a \le bn+2b$$ notice that $$a \le bn+b$$ si true then we can remove it and get $$an\le bn+2b$$ there is where I was stuck Thank you for anyone who could help me prove it with induction or by a simpler manner

Your "more general" problem would not imply the result you want since $1 + 1/n \ge 1 + 1/(n+1)$ and also these quantities are not integers when $n > 1$. — Daniel Arreola, Aug 12 '23 at 14:09
Oh just noticed I made a mistake while typing, I meant $b\le a$ not the opposite. — Raid Zougari, Aug 12 '23 at 14:15
You can find two proofs without induction in the answers to this question: https://math.stackexchange.com/questions/515317/how-is-a-n-11-nn-monotonically-increasing-and-bounded-by-3/4452536#4452536 — Adam Rubinson, Aug 12 '23 at 14:18
Does this answer your question? Prove $\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$ by mathematical induction - found using an Approach0 search. Note there are also quite a few other questions, and answers, here about proving the same inequality, but not necessarily using induction, e.g., ... — John Omielan, Aug 12 '23 at 17:14
(cont.) Proving : $\bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n}$, Prove the inequality $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}$, Showing the exponential sequence is strictly increasing, Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing, Arithmetic and Geometric Mean Inequality, The function $f(n)=(1+1/n)^{n+1}$ is decreasing, etc. There are also related AoPS threads, e.g., ... — John Omielan, Aug 12 '23 at 17:20
Thank you everyone for your help ! It was indeed very helpful ! Have a nice day ! — Raid Zougari, Aug 12 '23 at 21:27

score 2 · Accepted Answer · answered Aug 12 '23 at 14:19

2

$$\frac{\left(1+{1\over 1+n}\right)^{n+1}}{\left(1+{1\over n}\right)^{n}}= \left(1-\frac{1}{(n+1)^2} \right)^{n+1}\cdot \frac{n+1}{n}>\left(1-\frac{1}{n+1} \right)\cdot \frac{n+1}{n}=1$$

answered Aug 12 '23 at 14:19

zkutch

13,410

Thank you so much that was the type of proof I was looking for concise, even if I'd wish you développed more but I think I can do it myself thank you so much again for your help ! – Raid Zougari Aug 12 '23 at 14:33
Glad to be helpful. If you think this is the answer to your question, then you can select it. – zkutch Aug 12 '23 at 15:42

score 1 · Answer 2 · answered Aug 12 '23 at 14:35

1

Alternative Proof

Consider $$f(x)=\left(1+\frac{1}{x}\right)^x$$ $$f'=\frac{\left(1+\frac{1}{x}\right)^x\left((x+1)\ln\left(1+\frac{1}{x}\right)-1\right)}{x+1}$$ $$f'(x)>0 \implies \\ x>0$$ Since all natural numbers are $\ge0$ it will be an increasing function on that domain.

Another way is to use the expansion of the above mentioned function.

answered Aug 12 '23 at 14:35

MathStackexchangeIsNotSoBad

4,716

1

That a very intresting approach to the problem thank you so much for your help ! – Raid Zougari Aug 12 '23 at 14:36

How can I prove $\left(1+{1\over n}\right)^{n}\le\left(1+{1\over 1+n}\right)^{n+1}$ via induction?

2 Answers2