1

I just wanna some help showing that the sequence $a_n = \left(\frac{n}{n+1}\right)^{n+1}$ is increasing, since it came up in some stuff I was doing and I'm not finding a quick solution.

I actually just need it to be bounded below by a positive number, but I tested a bunch of values and it's increasing so I'm pretty sure it is.

Winther
  • 24,478
violeta
  • 11
  • 1
    I think there's a clever AM-GM inequality that works here. Rewriting the inside as $1 - \frac{1}{n+1}$ and shifting variables to get $\left(1-\frac{1}{n} \right)^n$ should help. – user217285 Sep 26 '15 at 22:43
  • Consider the function $f(x) = \left(\frac{x}{1+x}\right)^{x+1}$. If $f(x)$ is increasing so is $\log f(x)$ so try to show that $\frac{d\log f(x)}{dx} \geq 0$. The inequaity $\log(1+x) \leq x$ might be useful for this purpose. – Winther Sep 26 '15 at 22:47
  • 2
    Here's one way. – David Mitra Sep 26 '15 at 22:49
  • $$a_n = \left(\frac{n}{n+1}\right)^{n+1} = \left(1 - \frac{1}{n+1}\right)^{n+1}$$ Substitute $n+1 = u$ $$ = \left(1 - \frac{1}{u}\right)^{u}$$
    The limit as $n$ approaches infinity is $\frac{1}{e}$. Can this be used?     – Brevan Ellefsen Sep 26 '15 at 23:18

3 Answers3

5

We need only use Bernoulli's inequality and some straightforward arithmetic. To that end, we have

$$\begin{align} \frac{a_{n+1}}{a_{n}}&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\,\frac{n}{n+1}\\\\ &= \left(1+\frac{1}{n(n+2)}\right)^{n+2}\frac{n}{n+1}\\\\ &\ge \left(1+\frac1n\right)\frac{1}{1+\frac1n}\\\\ &=1 \end{align}$$

And we are done.

Mark Viola
  • 179,405
1

There is indeed an easy way to prove this by using AM-GM inequality for the following $n+2$ numbers: $$1,\frac{n}{n+1},\dots,\frac{n}{n+1},$$ where the fraction $\frac{n}{n+1}$ is repeated $n+1$ times. These numbers have as sum $n+1$. By AM-GM we get: $$\frac{n+1}{n+2}>\sqrt[n+2]{\left(\frac{n}{n+1}\right)^{n+1}},$$ which is obviously equivalent with the sequence $(a_n)$ being increasing. The inequality is strict since $1\ne \frac{n}{n+1}$.

JohnnyC
  • 1,220
0

I don't know if the process is appropriate in this case or if my methods are rigorus, but I think this can be done via Induction. If you can prove each term is greater than the previous than it follows that the sequence is increasing.

Let $P(n) : a_n \geq a_{n-1}$ or equivalently $P(n): (\frac{n}{n + 1})^{n+1} \geq (\frac{n - 1}{n})^n $ for $n \in \mathbb{Z}^+ | n \geq 1$.

First check the base case $P(1)$ \begin{align} (\frac{1}{2})^{2} > (\frac{0}{1})^1 \end{align} Therefore $P(1)$ holds. Then assume $P(k)$ holds. Now we must show that $P(k + 1)$ holds. Now from the inductive step it follows that \begin{align} (\frac{k}{k + 1})^{k+1} \geq (\frac{k - 1}{k})^k \end{align} Recall that if $a \geq b$ and $c \geq d$ then $ac \geq bd$. If we can prove that $\frac{k}{k+1} > \frac{k - 1}{k}$ $\forall k \geq 1$ then the proof is complete. I will prove it from the fact that $\frac{1}{x-1} > \frac{1}{x} \forall x \geq 2$ \begin{align} \frac{1}{x-1} &> \frac{1}{x}\\ 1-\frac{1}{x - 1} &< 1 - \frac{1}{x}\\ 1-\frac{1}{x-1} &< \frac{x - 1}{x}\\ \end{align} Let $k = x - 1$, therefore $k \geq 1$: \begin{align} 1 - \frac{1}{k} &< \frac{k}{k+1}\\ \frac{k - 1}{k} &< \frac{k}{k+1}\\ \end{align} Now that this has been proved, we can use the multiplicative principle and the inductive step \begin{align} (\frac{k}{k + 1})^{k+1} \cdot \frac{k}{k+1} &\geq (\frac{k - 1}{k})^k \cdot \frac{k - 1}{k}\\ (\frac{k}{k + 1})^{k+2} &\geq (\frac{k - 1}{k})^{k + 1}\\ \end{align} Therefore $P(k+1)$ holds if $P(k)$ does. From the principle of mathematical induction it follows that $P(n)$ holds for $n \in \mathbb{Z}^+ | n \geq 1$ holds as well. From this it follows that the sequence must be increasing.

Jeevan Devaranjan
  • 2,575