Convergence of the random harmonic series $\sum_{n=1}^{\infty}\frac{X_{n}}{n}$ Let $(X_{n})_{n \in \mathbb{N}}$ be independent with Rademacher distribution:

\begin{equation}
\mathbb{P}(X_{n} = -1) = \frac{1}{2} = \mathbb{P}(X_{n} = 1).
\end{equation}

I have to investigate

\begin{equation}
\sum_{n=1}^{\infty}\frac{X_{n}}{n}
\end{equation}
for convergence. It was given in a textbook and I'm very interested in the solution. It is something between the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ and the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$, but I know the sign changes randomly.
Define for $N \in \mathbb{N}$
\begin{equation}
Y_{N} = \sum_{n=1}^{N}X_{n}
\end{equation}
Now
\begin{equation}
\mathbb{E}(\frac{1}{n}X_{n}) = 0.
\end{equation}
Therefore
\begin{equation}
\mathbb{E}(Y_{N}) = 0.
\end{equation}
Also
\begin{align}
var(Y_{N}) &amp;= var(\sum_{n=1}^{N}X_{n})\\ 
&amp;= \sum_{n=1}^{N}\frac{1}{n^{2}}var(X_{n})\\
&amp;=\sum_{n=1}^{N}\frac{1}{n^{2}}\\
&amp;\leq\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}.
\end{align}

For $M &lt; N$ we have
\begin{align}
\mathbb{E}[|Y_{N}-Y_{M}|^{2}] &amp;= \mathbb{E}[|\sum_{n=M+1}^{N}\frac{1}{n}X_{n}|^{2}]\\
&amp;=var(\sum_{n=M+1}^{N}\frac{1}{n}X_{n})\\
&amp;=\sum_{n=M+1}^{N}\frac{1}{n^{2}} \xrightarrow{N,M \to \infty} 0,
\end{align}
i.e. $(Y_{N})_{N \in \mathbb{N}}$ is a cauchy-sequence with respect to the $\mathcal{L}^{2}$-norm. Now the Riesz–Fischer theorem says, that $\mathcal{L}^{2}$ is a banach-space. Especially every cauchy sequence in $\mathcal{L}^{2}$ converges. Now $\mathcal{L}^{2}$-convergence implies onvergence in probability and distribution. We also have almost sure convergence, since if $(X_{n})_{n \in \mathbb{N}}$ are independet random variables with $S_{n} = \sum_{i=1}^{n}X_{i} \xrightarrow{p} S$ then $S_{n} \rightarrow S$ almost surely.
Use the following theorem about sums of zero-mean independent variables in $\mathcal L^2$ from David Williams' Probability with Martingales Chapter 12.2.

Let $\{X_n\}_{n \in \mathbb N}$ be a sequence of random variables. We
have $$\sum_n X_n &lt; \infty \ \text{a.s.}.$$ if the following 4
conditions hold:

$$\{X_n\}_{n \in \mathbb N} \text{is independent.}$$

$$E[X_n] = 0 \ \forall n \in \mathbb N$$

$$\sigma_n^2 := Var[X_n] &lt; \infty \ \forall n \in \mathbb N$$

$$\sum_{n \in \mathbb N} \sigma_n^2 &lt; \infty$$



In our case:

$$\{\frac{X_n}{n}\}_{n \in \mathbb N} \ \text{is independent} \ \because \{X_n\}_{n \in \mathbb N} \text{is independent.}$$

$$E[\frac{X_n}{n}]=\frac{1}{n}E[X_n] = \frac{1}{n} 0 = 0 \ \forall n \in \mathbb N$$

$$Var[\frac{X_n}{n}] = \frac{1}{n^2} Var[\frac{X_n}{1}] = \frac{1}{n^2} (1)  = \frac{1}{n^2} &lt; \infty \ \forall n \in \mathbb N$$

$$\sum_{n \in \mathbb N} Var[\frac{X_n}{n}] = \sum_{n \in \mathbb N} \frac{1}{n^2} &lt; \infty \ \text{by integral test}$$


$\therefore,$ by theorem in Chapter 12.2 of David Williams' Probability with Martingales, $$\sum_{n \in \mathbb N} \frac{X_n}{n} &lt; \infty \ \text{a.s.}.$$ QED