It's well known that $\displaystyle\underset{n\geq1}\sum\frac{1}{n^2}=\frac{\pi^2}{6},$ which is less than $2$, hence $\displaystyle\underset{n\geq2}\sum\frac{1}{n^2}<1$. However, this inequality does not suffice to prove that the sequence of squares which sides are $\displaystyle\frac{1}{n}$ can be inscribed into the unit square, as in fact doesn't happen for the case of two squares with side $\displaystyle\frac{1}{2}+\varepsilon$, for which hold of course that the sum of their areas is less than $1$ for sufficiently small $\varepsilon>0$.
My question is: can this be proven to be true? I heard that the answer is yes, but my attempts arrived at most to describe a possible way to insert such squares (using the convergence of $\displaystyle\sum\frac{1}{(k+1)^n}=\frac{1}{k}$) as shown in the picture below. I really don't known if this is a good approach to the problem, it's just the nicest way I found to put them in and I was hoping that it works.
Thanks in advance :)
$\textbf{EDIT}$ As answered by JimN the solution is actually easy, though I continue to find it really smart! Arise then naturally another question: does the algorithm explained above converges remaining inside of the unit square?
Look at your horizontal lengths.. they form: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....$ so that pattern can continue on indefinitely as that summation will never reach 1 (or will reach 1 in the infinite limit).
Then looking vertically, you can create columns with the boxes sequentially, with 2 boxes in the column with $\frac{1}{2}$ and 4 boxes on the column with $\frac{1}{4}$ and so on. Then the column heights are:
$\frac{1}{2} + \frac{1}{3} < 1$
$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < 4\times \frac{1}{4} = 1$
Your $\frac{1}{9}$ should better be placed on top of your $\frac{1}{8}$ box, creating a column:
$\frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \ldots \frac{1}{15} < 8\times \frac{1}{8} = 1$
and so on, so each column fits on your square as well.