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?

  • $\begingroup$ Notice that $(n+1)/n > 1$ so this sequence is increasing. $\endgroup$ – CyclotomicField Apr 25 '20 at 13:19
  • 2
    $\begingroup$ @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. $\endgroup$ – Jean-Claude Arbaut Apr 25 '20 at 13:20
  • $\begingroup$ @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. $\endgroup$ – CyclotomicField Apr 25 '20 at 13:31
  • $\begingroup$ @Jean-ClaudeArbaut yes that's why they're called mistakes, because they're wrong. $\endgroup$ – CyclotomicField Apr 25 '20 at 13:38
  • $\begingroup$ Also: math.stackexchange.com/q/121076/42969 $\endgroup$ – Martin R Apr 25 '20 at 14:16



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., $h(y)<h(0)=0$. 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 $n$

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.