Three inequalities for number $e$: $\frac{e}{2n+2}

Question

I have to prove the following inequalities for number $e$:

$$\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}<\frac3{2n}$$

Here's what I have tried:

$\textbf{First inequality:}$ $\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n$

I use that $\log{\left(\frac{n+1}n\right)}=\int_n^{n+1}\frac1xdx\leq \frac12\left(\frac1n+\frac1{n+1}\right)$.

Operating we have: $n\log{\left(\frac{n+1}n\right)}\leq 1-\frac1{2n+2}$

Exponentiating and using $\frac1{1+x}\geq e^{-x}$, we have:

$$\frac{e}{2n+3}\leq e-\left(1+\frac1n\right)^n$$

This is similar to what i need to prove, but not the same...

$\textbf{Second inequality:}$ $e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}$

I don't know even how to start for this inequality.

$\textbf{Third inequality:}$ $\frac{e}{2n+1}<\frac3{2n}$

I think this is trivial as $e<3$ and $2n<2n+1$ (correct me if I am wrong).

Answer 1

Update: The book I mentioned below provides more than what you need. Here I give only the things you need to prove the two inequalities.

Step 1: We prove $a_n=\left(1+\frac 1n \right)^{n+\frac 12}$ is decreasing. You can check out this post Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? but the book's method is different and simpler.

Note that $$\log \frac{1+x}{1-x} = 2 \left( \frac x1 + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right)$$

Let $x=\frac{1}{2n+1}$ then $$ \log a_n = \left(n+\frac 12\right)\log \left(1+\frac 1n \right) = (2n+1)\left(\frac{1}{2n+1}+\frac{1}{3(2n+1)^3}+\frac{1}{5(2n+1)^5}+ \cdots \right)\\ =1+\frac{1}{3(2n+1)^2}+\frac{1}{5(2n+1)^4}+ \cdots $$ Therefore $a_n$ is (strictly) decreasing.

Step 2: We prove $b_n=\left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right)$ is monotone decreasing.

Note that $$b_n = a_n \frac{1+\frac{1}{2n}}{\left(1+\frac 1n\right)^{\frac 12}} = a_n \left( \frac{(1+\frac{1}{2n})^2}{1+\frac 1n}\right)^{\frac 12} = a_n \left(1+\frac{1}{4n(n+1)} \right)^{\frac 12}$$

Since $a_n$ and $1+\frac{1}{4n(n+1)}$ are both strictly decreasing, so is $b_n$.

Step 3: Now we go back to prove $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}$$

The first inequality $ \iff \left( 1+\frac 1n\right)^{n+1} < e\left( 1+\frac{1}{2n}\right)\tag 1$

Denote $f(x)=x+x \log \left( 1+\frac x2 \right)-(1+x)\log(1+x), 0 < x \leqslant \frac 1n.$

$$f'(x)=\frac{x}{x+2} - \log \frac{1+x}{1+\frac x2}>\frac{x}{x+2}-\frac{1+x}{1+\frac x2}+1=0, f(0)=0.$$

Therefore $$f\left(\frac 1n\right) >0 \implies (1).$$

The second inequality $$\iff e < \left(1+\frac 1n\right)^n \left(1+\frac{1}{2n}\right) \tag 2$$

From step 2, $\forall m>n$ $$ \left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2n+2}\right) \geqslant \left(1+\frac 1m\right)^m\left(1+\frac{1}{2m}\right)\\ \implies \left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2{n+1}}\right) \geqslant e, m\to \infty. \blacksquare $$

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These inequalities are not trivial. They are a series of problems from Problems and Theorems in Analysis I by Polya and Szego.

167. We define $$x_n = y_n e^{-\frac{1}{12n}}, y_n = n! n^{-n-\frac 12}e^n, n=1,2,3,\ldots$$

Then each interval $(x_k, y_k), k=1, 2, 3, \ldots, $ contains the interval $(x_{k+1}, y_{k+1})$ as a subinterval.

168. The sequence $$a_n=\left(1+\frac 1n \right)^{n+p}, n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $p\geqslant \frac 12$.

169. The sequence $$a_n=\left(1+\frac 1n\right)^n\left(1+\frac xn\right), n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $x\geqslant \frac 12.$

170. Let $n$ be a positive integer. Then we have $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}.$$