Find maximal a and minimal b in (1+1/n)^(n+a)≤e≤(1+1/n)^(n+b)

[lfdahl]

What is the maximal $a$ and the minimal $b$, such that:

$$\left(1+\frac{1}{n}\right)^{n+a} \le e \le \left(1+\frac{1}{n}\right)^{n+b} $$

holds for all natural numbers, $n$?

 

[Greg]

What is the maximal $a$ and the minimal $b$, such that:

$$\left(1+\frac{1}{n}\right)^{n+a} \le e \le \left(1+\frac{1}{n}\right)^{n+b} $$

holds for all natural numbers, $n$?

Spoiler

\(\displaystyle a=0,\quad b=\log_2(e)-1\)

 

[lfdahl]

Spoiler

\(\displaystyle a=0,\quad b=\log_2(e)-1\)

Spoiler

Hi, greg1313! Your answer is not correct. :(

 

[I like Serena]

Spoiler

I've been puzzling about this and by now I believe that the solution is $a=\frac 1{\ln 2} - 1=\log_2 e -1$ and $b=\frac 12$.
However, I can't quite prove it yet.
Either way, $a$ follows since it means that for $n=1$ it's an equality, and for higher $n$ it's below $e$.
We can find this $a$ by setting $(1+\frac 11)^{1+a}=e$,
$b=\frac 12$ is where the graph switches from an ascending graph for sufficiently high $n$ to a descending graph.
More generally, the graph changes from one with an intersection with $y=e$ for a positive value of $n$ to one with a negative value of $n$.

 

[lfdahl]

Spoiler

I've been puzzling about this and by now I believe that the solution is $a=\frac 1{\ln 2} - 1=\log_2 e -1$ and $b=\frac 12$.
However, I can't quite prove it yet.
Either way, $a$ follows since it means that for $n=1$ it's an equality, and for higher $n$ it's below $e$.
We can find this $a$ by setting $(1+\frac 11)^{1+a}=e$,
$b=\frac 12$ is where the graph switches from an ascending graph for sufficiently high $n$ to a descending graph.
More generally, the graph changes from one with an intersection with $y=e$ for a positive value of $n$ to one with a negative value of $n$.

Spoiler

Hi, I like Serena!
I can verify your presumption!;)
The way your graph varies, I don´t quite follow, but the values are correct.

 

[lfdahl]

Hint:

Spoiler

Analyse the behaviour of the function:

$$f(x) = \frac{1}{\ln (1+\frac{1}{x})}-x, \;\;\; x \ge 1.$$

 

[lfdahl]

Here´s the suggested solution:

Spoiler

The answer is $a_{max}=\frac{1}{\ln 2}-1$ and $b_{min}=\frac{1}{2}$. Consider the function:
\[f(x)=\frac{1}{\ln (1+\frac{1}{x})}-x\] with \[f(1)=\frac{1}{\ln (2)}-1>0.\]
Applying L´Hospitals rule twice, we have
\[\lim_{x\rightarrow \infty}f(x)=^{L´H} \lim_{x\rightarrow \infty} \frac{-\ln (1+\frac{1}{x})+\frac{1}{1+x}}{\frac{1}{1+x}-\frac{1}{x}}=^{LH} \lim_{x\rightarrow \infty} \frac{\frac{1}{x}-\frac{1}{1+x}-\frac{1}{(1+x)^2}}{\frac{1}{x^2}-\frac{1}{(1+x)^2}} \\\\ = \lim_{x\rightarrow \infty}\frac{\frac{1}{1+x}}{\frac{1}{x}+\frac{1}{1+x}} = \frac{1}{2}\]

Now, we prove that $f$ is increasing. In what follows, consider the three functions: $f,g,h : [1, \infty) \rightarrow \mathbb{R}$.
\[f'(x) = \frac{g(x)}{(\ln (1+\frac{1}{x}))^2}\] where \[g(x) = \frac{1}{x}-\frac{1}{1+x}-( \ln ( 1+1/x ) )^2\].
\[g'(x) = \left ( \frac{1}{x}-\frac{1}{1+x} \right )h(x)\] where \[h(x) = 2\ln (1+1/x)-\frac{1}{x}-\frac{1}{1+x}\]
and \[h'(x) = \left ( \frac{1}{x}-\frac{1}{1+x} \right )^2 > 0.\] Therefore $h$ is increasing. Now, $\lim_{x\rightarrow \infty}h(x) = 0$, and so $h < 0$. Therefore $g’(x) < 0$, $g$ is decreasing. Again:
$\lim_{x\rightarrow \infty}g(x) = 0$, and so $g(x) > 0$. Therefore $f’(x) > 0$, and so $f(x)$ is increasing.
Hence $f(1) \le f(x) < \frac{1}{2}$, so \[\ln \left ( 1+\frac{1}{x} \right )^{x+f(1)} \leq 1 \leq \ln \left ( 1+\frac{1}{x} \right )^{x+\frac{1}{2}}\].
Thus, we get for any $n \in \mathbb{N}$:

\[\left ( 1+\frac{1}{n} \right )^{n+\frac{1}{\ln 2}-1} \leq e < \left ( 1+\frac{1}{n} \right )^{n+\frac{1}{2}}.\]
Note, that $b_{min} = \frac{1}{2}$ is optimal but there is no $n$, such that the equality holds in the right side of the double inequality.