Convergence of r.v. sequence Say we have a sequence of independent r.v. $(X_n)^\infty_{n=1}$, we are given that $E[X_n]={\sqrt{n}}$
Is it true that the following holds a.s.?
$$\lim_{M \to \infty }\frac{1}{M}\sum^M_{n=1} X_n = 0 $$
 A: Warning: This post answers the original question:
$$\sum_{n=1}^{\infty} \frac{1}{n}X_n \to 0\;a.s.?$$

The scaling coefficient $\frac{1}{n}$ merely defines a new sequence of r.v.s $Y_n$ where: $E[Y_n]=\frac{1}{n}E[X_n]=\frac{1}{n^{3/2}}$. We can now make statements about $\sum^\infty_{n=1} Y_n$.
Lets first look at the expected value of the sequence of partial sums $\sum^M_{n=1} Y_n:= Z_M$:
$$E\left[Z_M\right] = \sum^M_{n=1} E\left[Y_n\right] = \sum^M_{n=1}\frac{1}{n^{3/2}}$$
From basic calculus, we know that $E[Z_M]$ remain finite as $M\to \infty$. However, we also know that $E[Z_M]>0\;\;\forall M \in \mathbb{N^+}$. 
Let's assume that $E[|X_n|]$ exists so we can make statements about mean convergence of $Z_M$ (If not, then the sequence will not converge $a.s.$ or here(see last page for Kolmorogov's SLLN).
Given the above, $Z_M$ converges in mean to an $a.s.$ constant random variable $K>0$; therefore it cannot approach zero $a.s.$ In fact, it won't even approach $0$ in probability.
We can see this via Markov's Inequality:
$$P(|Z_M-K|>\epsilon)\leq \frac{E[|Z_M-K|]}{\epsilon}$$
However, $Z_M \xrightarrow{1} K$, therefore, the RHS of the above inequality converges to $0$, which implies that $Z_M \xrightarrow{P} K \neq 0 \implies Z_M \nrightarrow 0 \text{ a.s.}$
A: I decided to keep the previous answer since it was already accepted. But here's the answer to the revised question (as currently posted).

As with my previous post, without loss of generality, we can assume $E[|X_n|]<\infty$, then we can state that:
$$X_n \xrightarrow{1} 0$$
This implies $X_n \xrightarrow{P} 0$ by similar augments from my previous post.
Unfortunately, we cannot go further without knowing something about your series $X_n$. These are codified in Kolmogorov's Three Series Theorem for some $A>0$ and independent random variables $(X_n)_{n \in \mathbb{N}}$


*

*$\sum_{n=1}^{\infty} P(|X_n| \geq A)$ converges

*If $Y_n:=X_n 1_{|X_n|\geq A}$ then $\sum_{n=1}^{\infty} E[Y_n]$ converges.

*$\sum_{n=1}^{\infty} Var(Y_n)$ converges



Some other ideas:
Lets define a new sequence from your definition: $Z_M := \frac{1}{M}\sum^M_{n=1} X_n$. Then, we know that:
$$E[Z_M] = \sum^M_{n=1} \frac{1}{M\sqrt{n}}$$
We need to know if this converges to $0$ as $M\to \infty$. First of all, we know that:
$$M>1 \implies 0<\sum^M_{n=1} \frac{1}{\sqrt{n}} < \sum^M_{n=1} 1=M \implies \frac{\sum^M_{n=1} \frac{1}{\sqrt{n}}}{M} < 1$$
So, we've shown that:
$$\lim_{M\to \infty} E[Z_M] < \infty$$
The question is how fast does this converge?
If we look at the limit of the partial sums of differences:
$$\lim_{M \to \infty}\sum^M_{n=1} \left\{1-\frac{1}{\sqrt{n}}\right\} =\infty \implies E[Z_M] = o(M) \implies E[Z_M] \to 0$$
Note: We can also see this by noting that $n \leq M$ so that the $\frac{1}{M\sqrt{n}} \leq \frac{1}{n^{3/2}}$, whose sum converges by the $p-series$ theorem.
So, we've shown that $Z_n \xrightarrow{1} 0$ which implies $Z_n \xrightarrow{P} 0$...but..we need more info to get to strong convergence (see the three-series theorem)...unfortunately, your $Z_M$ are not longer independent as well...so yet more issues.
