# Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$?

Why is it that $$\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},$$ for any natural numbers $$m, n$$?

I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $$m=n$$ but I am not sure if there is a rigorous proof for other cases?

Both sequences $$a_n=\left(1+\frac{1}{n}\right)^n, \quad b_n=\left(1+\frac{1}{n}\right)^{n+1}$$ are monotonic. In particular, $$a_n$$ is increasing and $$b_n$$ is decreasing.

Hence, if $$m,n\in \mathbb N$$ and $$k=$$max$$\{m,n\}$$, then $$a_n\le a_k\le b_k\le b_m.$$

Why are these sequences monotonic?

We have $$\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}n^n}{(n+1)^{2n+1}}=\frac{n+2}{n+1}\cdot\left(\frac{n^2+2n}{(n+1)^2}\right)^n \\ =\frac{n+2}{n+1}\cdot\left(1-\frac{1}{(n+1)^2}\right)^n \ge \frac{n+2}{n+1}\cdot\left(1-\frac{n}{(n+1)^2}\right)=\frac{(n+2)(n^2+n+1)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1.$$ In a similar fashion we get that $$b_n$$ is decreasing: $$\frac{b_n}{b_{n+1}}=\frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}}=\frac{(n+1)^{2n+3}}{n^{n+1}(n+2)^{n+2}}=\frac{n+1}{n+2}\cdot \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1} \\ = \frac{n+1}{n+2}\cdot \left(1+\frac{1}{n^2+2n}\right)^{n+1}\ge \frac{n+1}{n+2}\cdot \left(1+\frac{n+1}{n^2+2n}\right) =\frac{(n+1)(n^2+3n+1)}{(n+2)(n^2+2n)}=\frac{n^3+4n^2+4n+1}{n^3+4n^2+4n}>1.$$

• S. Symliris Interesting! – Z Ahmed Jan 17 at 8:43
• Yiorgos.Nice answer. To show that $b_k$ is decreasing is a bit tricky. math.stackexchange.com/questions/2071492/…. If this is proven, and knowing that $\lim a_n=e=\lim b_m$ $a_n\le e \le b_m$, the answer is obvious. – Peter Szilas Jan 17 at 9:38

The inequality holding for all naturals implies that

$$\left(1+\frac{1}{n}\right)^n < L\le U < \left(1+\frac{1}{m}\right)^{m+1},$$

where $$U=\inf_m\left(1+\dfrac{1}{m}\right)^{m+1}, L=\sup_n\left(1+\dfrac{1}{n}\right)^n$$. There may not be any overlap between the values in the LHS and RHS.

Indeed for all real $$x>0$$, $$\left(1+\frac1x\right)^x and $$\left(1+\frac1x\right)^{x+1}>e.$$

Both functions are monotonic and their limit $$x\to\infty$$ is $$e$$.

Monotonicity follows from the inequalities

$$\frac1{x+1}<\log\left(1+\frac1x\right)<\frac1x$$ which imply positive and negative derivatives respectively. These inequalities are supported by inequalities between the derivatives

$$-\frac1{(x+1)^2}>-\frac1{x(x+1)}>-\frac1{x^2}$$ and the common value $$0$$ at infinity.

• Sorry, @Yves Daoust I wanted to help. :) – Michael Rozenberg Jan 17 at 8:39
• @MichaelRozenberg: no problem, but the inequalities hold in all positives, and there was a typo in the directions. – Yves Daoust Jan 17 at 8:40
• Very interesting – Z Ahmed Jan 17 at 8:41

For a simple answer to why is $$\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1}$$? -- it is because of the inequality of geometric and arithmetic means. In particular

$$a_n = \big(1 + \frac{1}{n}\big)^n \lt \big(1 + \frac{1}{n+1}\big)^{n+1} = a_{n+1}$$
or equivalently, taking n+1 th roots
$$a_{n}^\frac{1}{n+1}= \big(1 + \frac{1}{n}\big)^\frac{n}{n+1} = \big(1 + \frac{1}{n}\big)^\frac{n}{n+1}\cdot 1^\frac{1}{n+1} \lt \frac{n}{n+1}\big(1 + \frac{1}{n}\big) + \frac{1}{n+1}\big(1\big) = 1 + \frac{1}{n+1} = a_{n+1}^\frac{1}{n+1}$$
by $$\text{GM}\leq \text{AM}$$ (which holds with equality iff all items in the geometric mean are identical)

and by the same manipulation we get $$c_r = \big(1 - \frac{1}{r}\big)^r \lt \big(1 + \frac{1}{r+1}\big)^{r+1} = c_{r+1}$$

so $$a_n$$ and $$c_r$$ are strictly monotone increasing

For the question posed: by monotone behavior of $$a_n$$, if $$n \leq m$$ then we automatically have
$$(1+\frac{1}{n})^n \leq (1+\frac{1}{m})^{m}\lt (1+\frac{1}{m})^{m}\cdot (1+\frac{1}{m}) = (1+\frac{1}{m})^{m+1}$$

it remains to consider the case of $$m \lt n$$. In this case we can divide out $$(1+\frac{1}{m})^{m+1}$$ and prove the equivalent

$$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} = \left(1+\frac{1}{n}\right)^n \left(\frac{m}{m+1}\right)^{m+1}=\left(1+\frac{1}{n}\right)^n \left(1+\frac{1}{m}\right)^{-(m+1)}\lt 1$$

but
$$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \leq \left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{n}\lt 1$$
by the monotone behavior of $$c_r$$ and $$\text{GM}\leq \text{AM}$$ which proves the claim

note: if you don't like splitting this into 2 cases, then the claim is proven directly by considering monotone behavior in $$a_n$$ and $$c_r$$, and for any $$m,n$$ select natural number $$k \gt \max(m+1,n)$$ and the direct proof is

$$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \lt \left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$$

and of course
$$\left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$$ is equivalent to proving the claim with 2kth roots of each side but

$$\left(1+\frac{1}{k}\right)^\frac{1}{2} \left(1-\frac{1}{k}\right)^\frac{1}{2}\lt \frac{1}{2}\left(1+\frac{1}{k}\right)+ \frac{1}{2}\left(1-\frac{1}{k}\right) = 1$$ by $$\text{GM}\leq \text{AM}$$