Q.Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.

[!TLDR] In an open pipe the fundamental has $L=\lambda/2$ (antinode-node-antinode); each higher mode adds one more half-wavelength, so $L=(p+1)\lambda/2$ for every integer p, giving $n_p=(p+1)n$ -- ALL harmonics …

A pipe open at BOTH ends must have an ANTINODE at each open end (section 6.7.3), since the air is comparatively free to move at both. The simplest (fundamental) pattern satisfying antinodes at BOTH ends has just one node in between, fitting exactly HALF a wavelength into the air column: $L=\lambda/2$, giving $n=v/(2L)$ -- the first harmonic. Unlike the closed-pipe case, here EVERY successive mode can be built by adding just ONE more half-wavelength segment (one more antinode-node-antinode unit) at a time, because both boundary conditions (antinode, antinode) are the SAME at each end -- there is no asymmetric constraint forcing the increment to be in steps of two, as there was for the closed pipe. So the allowed lengths run $L=\dfrac{\lambda}{2},\,\lambda_1,\,\dfrac{3\lambda_2}{2},\,2\lambda_3,\ldots$ -- EVERY positive-integer number of half-wavelengths is allowed, $L=(p+1)\dfrac{\lambda_p}{2}$ for $p=0,1,2,3,\ldots$. Since $v=n_p\lambda_p$ and $\lambda_p=\dfrac{2L}{p+1}$, the corresponding frequency is $n_p=\dfrac{v}{\lambda_p}=\dfrac{(p+1)v}{2L}=(p+1)n$, using $n=v/(2L)$ for the fundamental. B …

Apply the antinode-at-both-open-ends boundary conditions to find that any integer number of half-wavelengths can fit the air column, then convert thi …

Carrying over the closed-pipe reasoning (that the increment must be TWO quarter-wavelengths at a time) into the open-pipe case, missing that the symmetric antinode-antinode boundary he …