An inequality about e [closed]

Question

I happen to encounter the following inequality which I need to prove: $$\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right)>1,$$ for $k\in\mathbb{Z}^{+}$.

My current idea is to expand $$k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)=k+\frac{1}{2}+\frac{1}{24k}+O\left(\frac{1}{k^2}\right).$$

It seems like $$\left(1+\frac{1}{k}\right)^{k+\frac{1}{2}}>e$$ is always true? But not sure how can I proceed with the remaining $O\left(\frac{1}{k^2}\right)$ terms.

Answer 1

We can actually prove this directly for $k \ge 7$ and it should be easy to check it for $k<7$

We note that for $x < 1$ we have $x/2-x^2/3+x^3/4-x^4/5... \ge x/2-x^2/3$ (series converges absolutely and grouping in pairs the remainder is positive).

For $y<1$ we also have $1-e^{-y} \ge y-y^2/2$ again by using the Taylor series and grouping

So with $x=1/k, k \ge 2$ we have, $$1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k=1-e^{k\log(1+1/k)-1}=1-e^{-x/2+x^2/3-x^3/4+x^4/5 \ldots} \ge 1-e^{-x/2+x^2/3}$$

Also $$1-e^{-x/2+x^2/3} \ge x/2-x^2/3-(x/2-x^2/3)^2/2=x/2-11x^2/24+x^3/6-x^4/18$$

So $(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right) \ge \frac{1}{2}+\frac{1}{24 k}-\frac{7}{24 k^2}+\frac{1}{9 k^3}-\frac{1}{18 k^4} > \frac{1}{2}$ for $k \ge 7$

Hence $\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right) > (k+\frac{1}{2})\log\left(1+\frac{1}{k}\right), k \ge 7$

But if $f(x)=(x+\frac{1}{2})\log\left(1+\frac{1}{x}\right), x>1$ we have $f'(x)=\log\left(1+\frac{1}{x}\right)-\frac{1}{2x}-\frac{1}{2(x+1)}$ and then $f''(x)=-\frac{1}{x(x+1)}+\frac{1}{2x^2}+\frac{1}{2(x+1)^2}> 0$ so $f'$ increasing hence $f'<0$ (it is $0$ at infinity), hence $f$ decreasing so $f(x) >\lim_{y\to \infty}f(y)=1$ so indeed $(k+\frac{1}{2})\log\left(1+\frac{1}{k}\right)>1$ and finally $$\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right)>1, k \ge 7$$