Show that $ (b_{k})$ is strictly decreasing

Question

Where, $b_{k} := (1+ \frac{1}{k})^{k+1}$

From a previous part I have that:

$\frac{b_{k+1}}{b_{k}} = (1+ \frac{1}{k+1}) (1 - \frac{1}{(k+1)^2})^{k+1}$

How can I use this to show that $(b_{k})$ is strictly decreasing for $n \ge 1$?

My approach has been to try and show that $\frac{b_{k+1}}{b_{k}} <1$

My best attempt has been:

$\frac{b_{k+1}}{b_{k}} = (\frac{k+2}{k+1})(\frac{(k+1)^2-1}{(k+1)^2})(\frac{(k+1)^2-1}{(k+1)^2})^k$

I'm not worried about the last part $(\frac{(k+1)^2-1}{(k+1)^2})^k$, as it is clearly < 1, and $(\frac{(k+1)^2-1}{(k+1)^2})$ is also < 1, but $(\frac{k+2}{k+1})$ is clearly > 1, so I am having trouble combining this last term with $(\frac{(k+1)^2-1}{(k+1)^2})$, to try and get something that is clearly < 1.

I've also tried to use induction on $k$, but it just gets quite messy and nothing obvious sticks out. Any advice would be greatly appreciated!

EDIT:

I've also tried to use the concept of Infimum and Supremum, but I see that the supremum of $(1 + \frac{1}{k+1})$ occurs when $k = 1$ But the supremum of $(\frac{(k+1)^2-1}{(k+1)^2})^{k+1}$ is $1$ as $k \rightarrow \infty$. (Although it's clear that the index $k$ is required to be the same for both terms. But the supremum of the two terms together is quite hard to determine as they go in opposite directions as $k$ increases. Which still does not give me anything obvious. Any help would be great! Thanks!

Answer 1

Continuing your idea:

$\frac{b_{k+1}}{b_{k}} = (1+ \frac{1}{k+1}) (1 - \frac{1}{(k+1)^2})^{k+1} \stackrel{\text{Bernoulli}}{\le} \left((1+ \frac{1}{(k+1)^2})(1-\frac{1}{(k+1)^2})\right)^{k+1}\stackrel{\text{GM-AM}}{\le} 1^{k+1}=1$

Answer 2

since Use AM-GM inequality we have $$(1+\dfrac{1}{k})^k=(1+\dfrac{1}{k})(1+\dfrac{1}{k})\cdots(1+\dfrac{1}{k})\cdot 1\le\left(\dfrac{k+1+\dfrac{k}{k}}{k+1}\right)^{k+1}=\left(1+\dfrac{1}{k+1}\right)^{k+1}$$

Answer 3

$$ \begin{align} \frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}} &=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\ &=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\ &=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\ &\gt\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\ &=1 \end{align} $$ Where the strict inequality above follows from Bernoulli's Inequality, an inductive proof of which is given at the end of this answer.

Answer 4

Test some numbers and you'll see that $\frac{b_{k+1}}{b_{k}}>0$ for all $k$. For instance, with $k=1$ you get $\frac{9}{8}$.

It would be best to look over the previous part again. I cannot post comments yet, so I had to post this as an answer. My apologies.

Answer 5

Let $f$ be a function from $\mathbb{R}$ to itself defined by$$f(x)=(x+1)^{1/x}$$ Use logarithmic differentiation to show $$f'(x)=\frac{(x+1)^{\frac{1}{x}-1}(x-(x+1)\log(x+1))}{x^2}$$ Prove that $f'(x)<0 $ for $x\ge0$ (it holds for all $x$ but in your case you are likely only concerned with $x=k\in \mathbb{Z}_{\ge 0}$). What can we conclude from here?