The sample standard deviation is biased

Idea. $S^2$ is unbiased for ${\sigma }^2=1$, but the square root is concave, so $S$ itself is too small on average. With $n=3$ the distribution of $S$ is simple enough to compute $\mathbb{E}[S]$ exactly.

Step 1: the distribution. $W=2S^2$ is chi-square with 2 degrees of freedom, which is exponential with mean 2: density $\frac{1}{2}{e}^{-w/2}$.

Step 2: the mean. $S=\sqrt{W/2}$, so $$\begin{array}{rl}\mathbb{E}[S]& =\frac{1}{\sqrt{2}}{\int }_0^{\infty }\sqrt{w}\frac{1}{2}{e}^{-w/2}dw\\ & =\frac{1}{\sqrt{2}}\cdot \sqrt{2}\Gamma \left(\frac{3}{2}\right)=\frac{\sqrt{\pi }}{2}.\end{array}$$ (Substituting $w=2u$ gives $\sqrt{2}{\int }_0^{\infty }\sqrt{u}{e}^{-u}du=\sqrt{2}\Gamma (\frac{3}{2})$.)

So $\mathbb{E}[S]=\frac{\sqrt{\pi }}{2}\approx 0.886$: the sample standard deviation underestimates $\sigma =1$ by about 11%.

The general rule. $\mathbb{E}[S]=c_4(n)\sigma$ with $c_4(n)<1$ and $c_4(n)\to 1$ as $n$ grows. By Jensen's inequality, $\mathbb{E}[S]\le \sqrt{\mathbb{E}[S^2]}=\sigma$ always; dividing by $n-1$ fixes the variance, not the standard deviation.