Prove that $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1$.

Question

Let $n\in\Bbb N$, then $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1.$

Notice that,

$$\begin{align*}\left (1+\dfrac 1 {2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n&=\sum_{k=0}^n \binom{n}{k} \left (\dfrac{1}{2n}\right )^k-\sum_{k=0}^n\binom{n}{k}\left (-\dfrac{1}{2n}\right )^k\\ &=\sum_{k=0}^n \binom{n}{k}\left (\dfrac{1}{2n}\right )^k[1-(-1)^k] \end{align*}$$

I don't know how to get from there, because if $ n $ falls into an even number, all that goes to 0, and it is not greater than or equal to 1, but if $ n $ is odd if it is true. How do I finish the problem?

Answer 1

You have the right idea. Combining each of the corresponding terms in the binomial expansions of the $2$ powers gives

$$\begin{equation}\begin{aligned} \left(1 + \frac{1}{2n}\right)^n - \left(1 - \frac{1}{2n}\right)^n & = \sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{2n}\right)^{k} - \sum_{k=0}^{n}\binom{n}{k}\left(\frac{-1}{2n}\right)^{k} \\ & = \sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{2n}\right)^{k}\left(1 - (-1)^k\right) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note $1 - (-1)^k$ is either $0$, for even $k$, or $2$, for odd $k$. Thus, each of the terms above is non-negative, so the sum is that of the positive terms. The first term (i.e., for $k = 0$) is $0$, but the second term, i.e., $k = 1$, becomes

$$\binom{n}{1}\left(\frac{1}{2n}\right)^{1}(2) = n\left(\frac{1}{2n}\right)(2) = 1 \tag{2}\label{eq2A}$$

This means the sum's total is $1$ plus any additional positive terms and, thus, must be $\ge 1$ (basically as W. Wongcharoenbhorn's question comment suggests).

Answer 2

At least for large values of $n$ $$a_n=\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n$$

$$b=\left (1+\dfrac{1}{2n}\right )^n\implies \log(b)=n \log\left (1+\dfrac{1}{2n}\right )=\frac{1}{2}-\frac{1}{8 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$b=e^{\log(b)}=\sqrt e \left(1-\frac{1}{8 n}+\frac{19}{384 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ $$c=\left (1-\dfrac{1}{2n}\right )^n\implies \log(c)=n \log\left (1-\dfrac{1}{2n}\right )=-\frac{1}{2}-\frac{1}{8 n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$c=e^{\log(c)}=\frac{1}{\sqrt{e}} \left(1-\frac{1}{8 n}-\frac{13}{384 n^2}+O\left(\frac{1}{n^3}\right) \right )$$ $$a_n=2 \sinh \left(\frac{1}{2}\right)-\frac{\sinh \left(\frac{1}{2}\right)}{4 n}+\frac{13+19 e}{384 \sqrt{e} n^2}+O\left(\frac{1}{n^3}\right)$$ $$2 \sinh \left(\frac{1}{2}\right)-\frac{\sinh \left(\frac{1}{2}\right)}{4 n}\gt 1 \implies n > \frac{1}{8-4 \text{csch}\left(\frac{1}{2}\right)}\sim 3.08774$$

Computed exactly $$a_{10}=\frac{26372036801}{25600000000}=1.030158$$ while the truncated series gives $$a_{10}=\frac{37939 e-37907}{38400 \sqrt{e}}\sim 1.030184$$