Linear independence of $\begin{pmatrix}1\\ \vdots\\ \frac{1}{n}\end{pmatrix},\cdots,\begin{pmatrix}\frac{1}{n}\\ \vdots\\ \frac{1}{2n-1}\end{pmatrix}$

Question

I want to prove the linear independence of the set of n vectors

$$ \begin{pmatrix} 1 \\ \frac{1}{2} \\ \vdots \\ \frac{1}{n}\end{pmatrix}, \begin{pmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \vdots \\ \frac{1}{n+1}\end{pmatrix}, \cdots, \begin{pmatrix} \frac{1}{n} \\ \frac{1}{n+1} \\ \vdots \\ \frac{1}{2n-1}\end{pmatrix}$$

I am not sure if this statement is true, but it looks very intuitive. I tried writing $$ \alpha_1 \begin{pmatrix} 1 \\ \frac{1}{2} \\ \vdots \\ \frac{1}{n}\end{pmatrix} + \alpha_2 \begin{pmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \vdots \\ \frac{1}{n+1}\end{pmatrix} + \cdots + \alpha_n \begin{pmatrix} \frac{1}{n} \\ \frac{1}{n+1} \\ \vdots \\ \frac{1}{2n-1}\end{pmatrix} = 0$$ and showing $\alpha_i=0$, but the denominators are too messy and I am stuck. A lot of thanks in advance.

You'll get a Hilbert matrix, which is invertible. It's just so badly conditioned. — Sean Roberson, Aug 06 '23 at 04:11
A good question, and quite tricky, answered before in many ways on this site, e.g. the first half of this answer https://math.stackexchange.com/questions/430060/ — Eric Nathan Stucky, Aug 06 '23 at 09:48
Stack the vectors in a matrix and check if its determinant not equal to zero. — CroCo, Aug 06 '23 at 14:33

Linear independence of $\begin{pmatrix}1\\ \vdots\\ \frac{1}{n}\end{pmatrix},\cdots,\begin{pmatrix}\frac{1}{n}\\ \vdots\\ \frac{1}{2n-1}\end{pmatrix}$

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