let $n\in N^{+}$ show that $$\dfrac{e}{(1+\dfrac{1}{n})^n}<1+\dfrac{1}{2n}$$
My try: $$\Longrightarrow e<(1+\dfrac{1}{n})^n(1+\dfrac{1}{2n})$$
so let $$f(x)=x\ln{(1+\dfrac{1}{x})}+\ln{(1+\dfrac{1}{2x})}-1,x\ge 1$$ then $$f'(x)=\ln{(\dfrac{1}{x}+1)}-\dfrac{1}{x+1}-\dfrac{1}{2x^2+x}$$ then I can't $f'(x)>0(<0)?$
Thank you
$$1+\frac1{2n}=\frac12\left(2+\frac1n\right)=\frac12+\frac12\left(1+\frac1n\right)$$
So we have $$ \left(1+\frac1n\right)^n\left(1+\frac1{2n}\right)=\frac{\left(1+\frac1n\right)^n+\left(1+\frac1n\right)^{n+1}}{2}\ge \left(1+\frac1n\right)^{n+1/2} $$ The last step is given by AM-GM.
Now, let's find when $\left(1+\frac1n\right)^{n+1/2}\ge e$. Taking the logarithm, we obtain $$\begin{align} \left(n+\frac12\right)\log\left(1+\frac1n\right)\ge 1 \end{align}$$ Note that $$-\log(1-x)=\int\frac1{1-x}\text dx=\int\sum_{k=0}^\infty x^k\text dx=\sum_{k=1}^\infty \frac{x^k}{k}$$ So $\log\left(1+\frac1n\right)=-\sum_{k=1}^\infty \frac{(-1)^k}{kn^k}$ and we have $$\begin{align} \left(n+\frac12\right)\left(\sum_{k=1}^\infty \frac{(-1)^{k+1}}{kn^k}\right)&\ge 1\\ \sum_{k=0}^\infty\frac{(-1)^k}{(k+1)n^k}+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2kn^k}&\ge 1\\ \sum_{k=1}^\infty (-1)^{k}\frac{k}{2k(k+1)}n^{-k}&\ge 0 \end{align}$$
Note that this is an alternating series with terms that have a monotonically decreasing absolute value, therefore it is bounded below and above by its terms. The first term is $0$, so we may conclude the sum is $\ge 0$ and so the inequality holds for all $n$.
My answer here shows this more generally.
follow your solution: since $$f'(x)=\ln{\dfrac{x+1}{x}}-\dfrac{1}{x+1}-\dfrac{1}{x(2x+1)}$$ note $$\lim_{x\to+\infty}f'(x)=0$$ and $$f''(x)=\dfrac{4x+1}{(2x^2+x)^2}-\dfrac{1}{x^2+x}+\dfrac{1}{(x+1)^2}=\dfrac{5x^2+5x+1}{(2x^2+x)^2(x+1)^2}>0$$ so $$f'(x)<f(+\infty)=0$$ so $$f(x)>f(+\infty)=0$$ By Done!
