On the lengths of the cycles of random permutations A random permutation of $n$ from $n$ is chosen. What is the mathematical expectation of the sum of squares of lengths of its cycles?

The exponential generating function of permutations by the sum of the squares of the lengths of its cycles is
$$ G(z, u) = 
\exp
\left(u z + \frac{1}{2} u^4 z^2 + \frac{1}{3} u^9 z^3 + \frac{1}{4} u^{16} z^4 \cdots\right)$$
which means that the probability generating function of the expectation that we are looking for is
$$ \left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1} \\= 
\left.
\exp
\left(u z + \frac{1}{2} u^4 z^2 + \frac{1}{3} u^9 z^3 + \frac{1}{4} u^{16} z^4 \cdots\right)
\left(z + 2 u^3 z^2 + 3 u^8 z^3 + 4 u^{15} z^4 \cdots\right)
\right|_{u=1} \\=
\exp\log\frac{1}{1-z}\times \frac{z}{(1-z)^2} = 
\frac{1}{1-z}\times \frac{z}{(1-z)^2} =
\frac{z}{(1-z)^3}.$$
The conclusion is that the expectation we are trying to compute is given by
$$[z^n]  \left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1} =
[z^n] \frac{z}{(1-z)^3} = \frac{1}{2} n (n+1).$$
