How can I prove inequality $$ (1+\frac{1}{2n})^n<2 $$ for $n\in\mathbb{N}$ by using induction proof method?
I have tried to write that: for $n=1$ it's true. So let's suppose that $(1+\frac{1}{2n})^n<2$. and now I don't know what do nect because I know that $(1+\frac{1}{2(n+1)})^{n+1} > (1+\frac{1}{2n})^n$ but I don't know how to show that $(1+\frac{1}{2(n+1)})^{n+1} < 2$. In this task I cannot use the definition of $e$ number.
You can prove by induction that $$(1-x_1)(1-x_2)\cdots (1-x_n)\geq 1-x_1-x_2-\ldots -x_n$$ for all real numbers $x_1,x_2,\ldots,x_n\in[0,1]$ (the equality holds, by the way, iff at least $n-1$ of $x_1,x_2,\ldots,x_n$ are $0$). Using that inequality, we have $$\frac{1}{\left(1+\frac1{2n}\right)^n}=\left(1-\frac{1}{2n+1}\right)^n\geq 1-\frac{n}{2n+1}>\frac{1}{2}.$$
We can use the binomial theorem as in the proof for
$$\left(1+\frac1n\right)^n<e$$
refer to:
or as an alternative
$$\left(1+\frac{1}{2n}\right)^n<2\iff n\log \left(1+\frac{1}{2n}\right)<\frac12<\log 2$$
Hint: Use Bernoulli's inequality (one from here for instance http://lkozma.net/inequalities_cheat_sheet/ineq.pdf)
It is enough to show that $\log(2)\geq \frac{1}{2}$, since
$$ 2^{1/n} = e^{\frac{\log 2}{n}} > 1+\frac{\log 2}{n} \stackrel{\color{red}{?}}{\geq} 1+\frac{1}{2n}.$$ is granted by the convexity of $e^x$. On the other hand $\log(2)>\frac{1}{2}$ is equivalent to $e<4$, and
$$ e = \sum_{k\geq 0}\frac{1}{k!}=\frac{8}{3}+\sum_{k\geq 4}\frac{1}{k!}<\frac{8}{3}+\frac{1}{6}\sum_{k\geq 4}\frac{1}{4^{k-3}}=\frac{49}{18}. $$ As an alternative, $$ 0 < \int_{0}^{1}x^3(1-x^3)e^{-x}\,dx = \frac{1158}{e}-426 $$ directly gives $e<\frac{193}{71}$. Or, more elementary: $$ \log(2)-\frac{1}{2}=\int_{0}^{1}\frac{dx}{1+x}-\int_{0}^{1}\frac{dx}{1+1}=\frac{1}{2}\int_{0}^{1}\frac{1-x}{1+x}\,dx > 0.$$
Yet another approach: for any $n\geq 1$, $2^{1/n}$ is the geometric mean of $\frac{n+1}{n},\frac{n+2}{n+1},\ldots,\frac{2n}{2n-1}$.
By the AM-GM inequality it follows that
$$\begin{eqnarray*} 2^{1/n} < \frac{1}{n}\sum_{k=0}^{n-1}\left(1+\frac{1}{n+k}\right)&=&1+\frac{1}{n^2}+\frac{1}{n}\sum_{k=1}^{n-1}\frac{1}{k+n}\\&<&1+\frac{1}{n^2}+\frac{1}{n}\sum_{k=1}^{n-1}\frac{1}{\sqrt{k+n}\sqrt{k+n-1}}\end{eqnarray*} $$
and by the Cauchy-Schwarz inequality (see also page 8 of my notes) we have
$$ \sum_{k=1}^{n-1}\frac{1}{\sqrt{k+n}\sqrt{k+n-1}}\leq \sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\left(\frac{1}{k+n-1}-\frac{1}{k+n}\right)}=\frac{n-1}{\sqrt{n(2n-1)}},$$
so $2^{1/n}< 1+ \frac{1}{n\sqrt{2}}+\frac{1}{n^2}$ and actually $2^x< 1+\frac{x}{\sqrt{2}}+x^2$ for any $x$ in a pointed neighbourhood of the origin. By replacing $x$ with $-x$ and reciprocating we get
$$ 2^x > \frac{1}{1-\frac{x}{\sqrt{2}}+x^2} $$
and
$$ 2^{1/n} > \frac{1}{1-\frac{1}{n\sqrt{2}}+\frac{1}{n^2}} $$
still is stronger than needed.
