# Prove $(\frac{n+1}{n})^{n+1}$ is decreasing [duplicate]

I managed to prove a similar fact: the following sequence is increasing: $$\left( \dfrac{n+1}{n}\right)^n$$, which means $$\left( \dfrac{n+1}{n}\right)^n < \left( \dfrac{(n+1)+1}{(n+1)}\right)^{n+1}$$.

All this took was some simple algebra and bernoulli's inequality. For the one in the title, I am not sure how.
My attempt was: $$\left(\frac{n+1}{n}\right)^n \left(\frac{n+1}{n}\right)\stackrel{?}{>}\left(\frac{(n+1)+1}{(n+1)}\right)^{n+1} \left(\frac{(n+1)+1}{(n+1)}\right)$$ But this doesn't help me. Do you have other ideas?

• Notice that $(n+1)/n > 1$ so this sequence is increasing. – CyclotomicField Apr 25 '20 at 13:19
• @TobyMak Really? You should try numerically then. The sequence is indeed decreasing. It's one of the classical ways to prove the limit of $(1+1/n)^n$ exists. – Jean-Claude Arbaut Apr 25 '20 at 13:20
• @Jean-ClaudeArbaut I have convinced myself many times that the limit of $(1 + 1/n)^n$ doesn't exist and this seems to be yet another case of the same mistake. – CyclotomicField Apr 25 '20 at 13:31
• @Jean-ClaudeArbaut yes that's why they're called mistakes, because they're wrong. – CyclotomicField Apr 25 '20 at 13:38
• – Martin R Apr 25 '20 at 14:16

$$\frac{\left(\frac{n+1}{n}\right)^{n+1}}{\left(\frac{n}{n-1}\right)^n}=\frac{(n+1)^{n+1}(n-1)^n}{n^{2n+1}}=\underbrace{\left(1+\frac{1}{n}\right)}_{\leqslant\left(1+\frac{1}{n^2}\right)^n}\left(1-\frac{1}{n^2}\right)^n\leqslant\left(1-\frac{1}{n^4}\right)^n<1.$$

Although this question is labeled precalculus, there is a calculus approach to show that this is decreasing.

Let's begin by considering the function $$f(x)=\left(\frac{x+1}{x}\right)^{x+1}.$$ If we can show that this function is decreasing, then we've shown that the original sequence is decreasing. Since the natural logarithm is an increasing function, we know that $$f(x)$$ is decreasing if and only if $$\ln(f(x))$$ is a decreasing function. So, we consider $$g(x)=\ln(f(x))=(x+1)\ln\left(\frac{x+1}{x}\right).$$

Now, we take the derivative of $$g$$ to get $$g'(x)=\ln\left(1+\frac{1}{x}\right)-\frac{1}{x}.$$ Now, it is enough to show that $$g'(x)<0$$ for all $$x$$. Consider the function $$h(y)=\ln(1+y)-y$$. If we can show that this function is negative for $$y>0$$, then, since $$g'(x)=h\left(\frac{1}{x}\right)$$, the result has been shown. Observe that $$h(0)=0$$ and consider $$h'(y)=\frac{1}{1+y}-1=\frac{-y}{1+y}.$$ For $$y>0$$, the derivative $$h'(y)<0$$, so $$h(y)$$ is decreasing, i.e., $$h(y). Therefore, we know that $$g'(x)<0$$ and so $$f(x)$$ is decreasing.

Easy approach with A.M$$\gt$$G.M

Let $$u_n=(\frac{n+1}{n})^{n+1}=\left(1+\frac{1}{n}\right)^{n+1}$$ Let us consider n+2 positive numbers $$(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),••••,(1-\frac{1}{n+1}),[(n+1)times]$$and $$1$$.

Applying A.M$$\gt$$G.M.,we have,

$$\frac{(n+1)(1-\frac{1}{n+1})+1}{n+2}\gt (1-\frac{1}{n+1})^{\frac{n+1}{n+2}}$$

Or,$$(\frac{n+1}{n+2})^{n+2}\gt (\frac{n}{n+1})^{n+1}$$

Or,$$(\frac{n+1}{n})^{n+1}\gt (\frac{n+2}{n+1})^{n+2}$$ That is $$u_n\gt u_{n+1}$$ for all $$n$$

• Syntax issues :-( . – Oscar Lanzi Apr 25 '20 at 13:53
• @Oscar Lanzi Sorry mate,now see,I have edited. – Nimu Basak Apr 25 '20 at 13:58