This post directly proves that
$$ P\left( \sum_{n=1}^{\infty} \frac{a_n}{n} = \infty \right) = 1$$
rather than study about different quantities so as to make a heuristic argument that helps convince that the result is plausible.
The method is
- Decompose the sum into a sequence of independent, disjoint partial sums with rapidly decreasing variance
- Find a lower bound on the probability that each partial sum is greater than $0.01$
- Find a lower bound on the probability that infinitely many of the partial sums are all greater than $0.01$ (and thus the sum diverges)
- Deduce from the form of the final result that the probability of divergence is $1$
Borrowing the methods of the other answer,
Let $X_n = a_n / n$ be the $n$-th term.
Define a new sequence of independent random variables
$$ S_n = \sum_{i=2^n}^{2^{n+1} - 1} X_n $$
so that the total sum is
$$ \sum_{i=1}^{\infty} \frac{a_i}{i} = \sum_{n=0}^{\infty} S_n $$
We have
$$ E[S_n] = \sum_{i=2^n}^{2^{n+1} - 1} E[X_n ] = \frac{1}{2} (H_{2^{n+1} - 1} - H_{2^n - 1} ) \approx \frac{1}{2} \ln(2)$$
where $H_n$ is the $n$-th harmonic number. We also have the variance
$$ Var[S_n] = \sum_{i=2^n}^{2^{n+1} - 1} \frac{1}{4i^2} < 2^{-n} $$
(we can prove a slightly stronger bound, but I'm lazy!)
Chebyshev's inequality then says that, for any positive $k$,
$$ P\left(\left|S_n - \frac{1}{2} \ln 2\right| < k 2^{-n/2} \right) \geq 1 - \frac{1}{k^2} $$
For sufficiently large $k$, I claim that
$$ \exp(-2/k^2) = 1 - \frac{2}{k^2} + \frac{2}{k^4} - \ldots \leq 1 - \frac{1}{k^2} $$
and therefore,
$$ P\left(\left|S_n - \frac{1}{2} \ln 2\right| < \frac{1}{4} \right) \geq 1 - \frac{1}{2^{n-4}} \geq \exp\left(-\frac{1}{2^{n-5}} \right) $$
These events are jointly independent, so we have
$$ P\left(\forall n \geq m: \left|S_n - \frac{1}{2} \ln 2\right| < \frac{1}{4} \right) \geq \exp\left(-\sum_{n = m}^{\infty} \frac{1}{2^{n-5}} \right) = \exp(-2^{6-m}) $$
In particular, this means
$$ P(\forall n \geq m: S_n > 0.01) \geq \exp(-2^{6-m}) $$
from which it follows
$$ P\left( \sum_{n=0}^{\infty} S_n = \infty \right) \geq \exp(-2^{6-m}) $$
by taking the limit as $m \to \infty$, we get
$$ P\left( \sum_{n=0}^{\infty} S_n = \infty \right) \geq 1 $$