11. Given $a_{n}=\frac{1}{n}$, the squares with side lengths $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ can be placed inside a square of side length $a$ without overlapping. Find the minimum value of $a$.
As shown in the figure, squares with side lengths $a_{1}=1, a_{2}=\frac{1}{2}$ can be placed inside a square with side length $a$ without overlapping, then $a \geqslant \frac{3}{2}$.
For $k \in \mathbf{N}^{*}, a_{2^{k}}+a_{2^{k}+1}+a_{2^{k}+2}+\cdots+a_{2^{k+1}-1}$
$$
\begin{array}{l}
=\frac{1}{2^{k}}+\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}-1} \\
<\underbrace{\frac{1}{2^{k}}+\frac{1}{2^{k}}+\frac{1}{2^{k}}+\cdots+\frac{1}{2^{k}}}_{2^{k} \uparrow}=1,
\end{array}
$$

Thus, squares with side lengths $a_{2^{k}}, a_{2^{k}+1}, a_{2^{k}+2}, \cdots, a_{2^{k+1}-1}$ can be placed inside a rectangle of length 1 and height $\frac{1}{2^{k}}$ without overlapping.
When $k=2$, squares with side lengths $a_{4}, a_{5}, a_{6}, a_{7}$ can be placed inside a rectangle of length 1 and height $\frac{1}{2^{2}}$ without overlapping; when $k=3$, squares with side lengths $a_{8}, a_{9}, \cdots, a_{15}$ can be placed inside a rectangle of length 1 and height $\frac{1}{2^{3}}$ without overlapping; $\cdots$; therefore, squares with side lengths $a_{4}, a_{5}, \cdots, a_{n}, \cdots$ can be placed inside a series of rectangles of length 1 and heights $\frac{1}{2^{2}}, \frac{1}{2^{3}}, \cdots$ without overlapping, and $\lim _{n \rightarrow+\infty}\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n}}\right)=\frac{\frac{1}{2^{2}}}{1-\frac{1}{2}}=\frac{1}{2}$, meaning that squares with side lengths $a_{4}, a_{5}, \cdots, a_{n}$, $\cdots$ can be placed inside a rectangle of length 1 and height $\frac{1}{2}$ without overlapping.
Furthermore, when $k=1$, squares with side lengths $a_{2}, a_{3}$ can be placed inside a rectangle of length 1 and height $\frac{1}{2}$ without overlapping, so squares with side lengths $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ can be placed inside a square of side length $\frac{3}{2}$ without overlapping.
In summary, the minimum value of $a$ is $\frac{3}{2}$.