I am working on the following sequences.

$$x_n=\left(1+\frac1n\right)^n \qquad z_n=\left(1-\frac1n\right)^n \qquad y_n=\left(1+\frac1n\right)^{n+1}$$

I am trying to prove that $x_n$ and $z_n$ are increasing sequences and $y_n$ is a decreasing sequence.

  • I proved that $x_n$ is increasing using the binomial theorem: as $n$ increases, the number of terms in my sum increases. Also, for $k\ge1$ and $n\ge k$ the $(k+1)th$ term in our sum is $$\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\left(\frac{1}{n}\right)^k$$ which can be rewritten as $$\frac{1}{k!}\left[\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}{n}\right)\right]\qquad(1)$$ Here, the product in $(1)$ increases as $n$ increases.

  • I am in trouble proving that $z_n$ and $y_n$ are decreasing. Could you suggest me an approach to this problem please?

Thank you.


Since I obtained an answer for $y_n$, I kept working on $z_n$. I took the continuous function $z(x)=(1-\frac1x)^x=\exp \left\{x\log(1-\frac1x)\right\}$. Then I derived it to obtain $$z'(x)=\exp\left\{x\log(1-\frac1x)\right\}\left[\log(1-\frac1x)+\frac{x}{(x-1)/x}\cdot\frac{1}{x^2}\right]$$ But $$z'(x)=\exp\left\{x\log(1-\frac1x)\right\}\left[\log(1-\frac1x)+\frac{1}{x-1}\right]$$

$e^a$ is always positive. Hence the sign of $z'(x)$ depends on what we have between square brackets, which is greater than zero (I don't know how to prove it; I used a plot) for $x\ge1$. Hence $z_n$ is increasing.

  • 4
    $\begingroup$ Hint: Write $z_n=\exp(v(n))$ and $y_n=\exp(u(n))$ for some functions $v$ and $u$ and show that $u$ and $v$ are decreasing, using their derivatives $u'$ and $v'$. (Alternatively, write $z_n=\exp(v(1/n))$ and $y_n=\exp(u(1/n))$ and show that $u$ and $v$ are increasing.) $\endgroup$
    – Did
    Sep 10, 2014 at 9:38
  • 1
    $\begingroup$ @Did what do you think of my try? $\endgroup$
    – Charlie
    Sep 10, 2014 at 13:40
  • 1
    $\begingroup$ for the middle part see math.stackexchange.com/a/764461/129458 $\endgroup$
    – OBDA
    Sep 10, 2014 at 14:03
  • 1
    $\begingroup$ @Charlie That it can be completed using the bound $\log(1+u)\leqslant u$ for $u=1/(x-1)$. $\endgroup$
    – Did
    Sep 10, 2014 at 14:40

3 Answers 3


For $y_n$ : Let $A=\frac{n^2-1}{n^2},B=\frac{n+1}{n}$. By AM-GM inequality, we have $$\begin{align}\sqrt[n+1]{A^nB}\lt\frac{nA+B}{n+1}&\iff A^nB\lt\left(\frac{nA+B}{n+1}\right)^{n+1}\\&\iff \left(\frac{n^2-1}{n^2}\right)^n\cdot \frac{n+1}{n}\lt \left\{\frac{1}{n+1}\left(\frac{n(n^2-1)}{n^2}+\frac{n+1}{n}\right)\right\}^{n+1}\\&\iff \left(\frac{n^2-1}{n^2}\right)^n\cdot \frac{n+1}{n}\lt 1^{n+1}=1\\&\iff \left(\frac{1+\frac 1n}{1+\frac{1}{n-1}}\right)^n\left(1+\frac 1n\right)\lt 1\\&\iff y_{n-1}\gt y_n.\end{align}$$

  • $\begingroup$ I can follow your creative steps. But I cannot see the answer to my question. To which question are you answering? And how did you think of AM-GM? $\endgroup$
    – Charlie
    Sep 10, 2014 at 9:58
  • $\begingroup$ @Charlie: See the last of my answer. I proved $y_{n-1}\gt y_n$, which means $y_n$ is a decreasing sequence. And I use AM-GM inequality just because I've known this proof. $\endgroup$
    – mathlove
    Sep 10, 2014 at 10:04

One approach would be to take the continuous function, say $y(x)=(1+\frac1x)^{x+1}$ and show that $y'(x)<0$ for large enough $x$. In this example,

$$ y'(x)=(1+\frac1x)^{x+1}\left(\log\left(1+\frac1x\right)-\frac1x\right) $$

But the function $\log(1+t)$ is convex, so it lies below its tangent at $t=1$, so $\log(1+t)\leq t$. This gives you that $y'(x)\leq0$ for all $x$ by putting $t=1/x$. This is sufficient to show that $y(x)$ is decreasing, and hence so is $y_n$. The others can be done similarly.


EDIT: it seems there is a mistake in one of the steps of this solution. I'm only leaving it up, because others might try a similar approach. See the comments below for the mistake.

For $z_n$, examine the ratio of consecutive terms in the sequence.

$$ \frac{z_{n+1}}{z_n}=\frac{\left( 1-\frac{1}{n+1}\right) ^{n+1}}{\left( 1-\frac{1}{n}\right) ^{n}}=\frac{\left( \frac{n}{n+1} \right)^{n+1}}{\left( \frac{n-1}{n} \right)^{n}} $$

But we know that

$$ n^2 > n^2-1 \Rightarrow \\ \Rightarrow n n > \left(n-1 \right) \left(n+1 \right) \Rightarrow \\ \Rightarrow \frac{n}{n+1} > \frac{n-1}{n} \Rightarrow \\ \Rightarrow \left(\frac{n}{n+1}\right)^{n+1} > \left(\frac{n-1}{n} \right)^n \Rightarrow \\ \Rightarrow \frac{\left( \frac{n}{n+1} \right)^{n+1}}{\left( \frac{n-1}{n} \right)^{n}} > 1 \Rightarrow \\ \Rightarrow z_{n+1} > z_n$$

Alternatively: $$ n+1>n \Rightarrow \\ \Rightarrow -\left(n+1\right)<-n \Rightarrow \\ \Rightarrow -\frac{1}{n+1} > -\frac{1}{n} \Rightarrow \\ \Rightarrow 1-\frac{1}{n+1} > 1-\frac{1}{n} \Rightarrow \\ \Rightarrow \left(1-\frac{1}{n+1} \right)^{n+1} > \left(1-\frac{1}{n} \right)^n\Rightarrow \\ \Rightarrow z_{n+1} > z_n $$

Both methods prove that $z_n$ is increasing which is the opposite of what you asked, but the methods seem pretty solid to me.

  • $\begingroup$ Unfortunately, the implication $$x>y\Rightarrow x^{n+1} > y^n,$$ which you use in both approaches in your answer for $$x=\frac{n}{n+1},\qquad y= \frac{n-1}{n},$$ does not hold in general. $\endgroup$
    – Did
    Sep 10, 2014 at 10:17
  • $\begingroup$ Now that you mention it I thought it was true for $x>y>0$ but on second inspection it seems to only hold for $x>y>1$ which is not the case here. $\endgroup$ Sep 10, 2014 at 10:34

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