Sequence of finite, positive and i.i.d random variables and limit of $\frac{S_{n+1}}{S_{n}}$ Let $(X_{n})_{n\in\mathbb{N}}$ be a sequence of finite, positive and i.i.d random variables and let's call $\mu:=E(X_{1})>0$ and $S_{n}:=\sum_{i=1}^{n}X_{i}$. We know that $\dfrac{S_{n}}{n}\longrightarrow \mu$ a.s (in particular, $S_{n}=n\dfrac{S_{n}}{n}\longrightarrow +\infty$ a.s).
Is it true that $\dfrac{S_{n+1}}{S_{n}}\longrightarrow 1$ a.s ?
Well, when $\mu<+\infty$, it's easy to see that it's true: $\dfrac{S_{n+1}}{S_{n}}=\dfrac{n+1}{n}\dfrac{\frac{S_{n+1}}{n+1}}{\frac{S_{n}}{n}}\longrightarrow 1$ a.s
My difficulty is the case  $\mu=+\infty$
 A: The problem is to determinate whether the sequence $(X_{n +1}/S_n)_{n\geqslant 1}$ converges to $0$ almost surely. One can see that the convergence in probability takes place:
$$\mathbb P\left\{\frac{X_{n+1}}{S_n}\gt \varepsilon      \right\}\leqslant \mathbb P\{X_1\gt n\sqrt \varepsilon\} +\mathbb P\left\{\frac{n}{S_n}\gt \sqrt \varepsilon      \right\}.$$
An other observation is that the random variable $L:=\limsup_{n\to \infty}X_{n +1}/S_n$ is almost surely constant, and possibly infinite. 
Here is an example where $L= +\infty$. Consider a strictly increasing sequence of integers $(n_j)_{j\geqslant 1}$ such that $n_{j +1}\geqslant j^2n_j$ for each $j$ (for example $n_j:=(j!)^2$). We consider an i.i.d. sequence $(X_n)_{n\geqslant 1}$ such that $\mathbb P(X_1=n_j)=1/j-1/(j+1)$ ($X_1$ takes only its values among the $n_j$'s). 
Notice that 
$$\left\{\frac{X_{j+1}} {S_j}\geqslant j     \right\} \supset 
\{X_{j+1}\geqslant n_{j+1}\}\cap\bigcap_{i=1}^j\{X_i\leqslant n_j\}=:A_{j+1},$$
hence it suffices to show that the event $A:=\limsup_{j\to \infty}A_j$ has a positive probability. Defining $\mathcal G_j:=\sigma(X_1,\dots,X_j)$, the martingale convergence theorem shows that 
$$\mu(A)=\mu\left\{\sum_{j=1}^{\infty}\mathbb E[\mathbf 1(A_{j+1})\mid \mathcal G_j  ] = +\infty   \right\}.$$
Notice that by independence, 
$$\mathbb E[\mathbf 1(A_{j+1})\mid \mathcal G_j  ]=\frac 1{j+1}\mathbf 1\left(\bigcap_{i=1}^j\{X_i\leqslant n_j\}\right),  
$$
hence defining 
$$B:=\left\{\sum_{j=N+1}^{2N}\frac 1j \mathbf 1\left(\bigcap_{i=1}^j\{X_i\leqslant n_j\}\right)\mbox{ does not converge to }0   \right\},$$
the proof will be complete if we show that $B$ has a positive probability. 
Notice that $B$ contains the event 
$$C:= \left\{\mathbf 1\left(\bigcap_{i=1}^{2N} \{X_i\leqslant n_{N+1} \}\right)\mbox{ does not converge to }0   \right\},$$
which is noting but 
$$\limsup_{N\to \infty}\bigcap_{i=1}^{2N} \{X_i\leqslant n_{N+1} \}.$$
Since for $N$ large enough, again by independence, 
$$\mu\left(\bigcap_{i=1}^{2N} \{X_i\leqslant n_{N+1} \}\right)=
\prod_{i=1}^{2N}\left(\sum_{l=1}^{N+1}\frac 1l-\frac 1{l+1}      \right)  =\left(1-\frac 1{N+2} \right)^{2N} \geqslant \frac{1}{2e^2},$$
we get that $\mu(C)\gt 0$, hence 
$$\mu\left(\limsup_{j\to \infty}\left\{\frac{X_{j+1}} {S_j}\geqslant j     \right\}   \right)\gt 0.$$
