# Showing almost sure divergence

Doing some exercises as preparation for an upcoming exam, but im sort of stuck at this exercise: \

Assume that $X_1,X_2,...$ is an i.i.d sequence, such that $X_1 \sim \mathcal{N} (\xi , \sigma^2)$, with $\xi>0$ . Define $$S_n=\sum_{k=1}^n \frac{X_k}{k}$$ Show that $X_n \to \infty$ almost surely as n tends to infinity.

Im not aware of any theorem that can ease my way of showing this quickly. But im trying something going this direction: $$X_n \stackrel{a.s.}{\to} \infty \iff \forall \varepsilon>0 : P(S_n>\varepsilon \quad evt.)=1 \iff \forall \varepsilon>0: P( S_n \leq \varepsilon \quad i.o.)=0$$ $$\Leftarrow \forall \varepsilon>0 : \sum_{k=1}^\infty P(S_n \leq \varepsilon) <\infty$$ And from here i dont know where to go, because i have no closed form expression of the above probabilities, or even any idea if it holds.

Any tips/tricks/solutions are welcome.

## 2 Answers

Let's begin with a result that has nothing to do with probability. If $(x_k)$ is a sequence of numbers where $\sum_{k=1}^n {x_k\over k}$ converges to a finite limit $\alpha$, then ${1\over n}\sum_{k=1}^n x_k\to 0$.

Proof: Note that $\sum_{k=1}^n {x_k\over k}\to\alpha$ implies that the Cesàro averages also converge to $\alpha$: $${1\over n}\sum_{j=2}^n \left( \sum_{k=1}^{j-1}{x_k\over k}\right)\to\alpha.$$ But since

$$\begin{eqnarray*} {1\over n}\sum_{j=2}^n \left( \sum_{k=1}^{j-1}{x_k\over k}\right) &=&{1\over n}\sum_{k=1}^n \left( \sum_{j=k+1}^n {x_k\over k}\right)\\[5pt] &=&\sum_{k=1}^n{x_k\over k}-{1\over n}\sum_{k=1}^n x_k. \end{eqnarray*}$$

we deduce that ${1\over n}\sum_{k=1}^n x_k\to 0$.

Now, let's put probability back into the picture. The law of large numbers gives ${1\over n}\sum_{k=1}^n X_k\to\xi>0$ and therefore $\sum_{k=1}^n {X_k\over k}$ cannot converge to a finite value.

• Thanks for the answer! I have never heard of Cesaro averages before, but they did the trick! – Martin Nov 3 '13 at 11:54
• Actually, this argument only shows that the limit cannot be finite but doesn't actually give convergence. This can be fixed, but frankly, I think you are better off with Lord Soth's solution. – user940 Nov 3 '13 at 15:04

Hint: $X_k/k \sim \mathcal{N}(\xi/k, \sigma^2/k^2)$. Then, $S_n \sim \mathcal{N}(\xi\sum_{k=1}^n\frac{1}{k},\sigma^2 \sum_{k=1}^n \frac{1}{k^2})$. The mean grows to infinity, but the variance is at most $\sigma^2 \frac{\pi^2}{6}$.

• Okay, in a heuristic argument it would be clear after stating the above. But how would you use this information, if a formal deduction is needed? – Martin Nov 3 '13 at 11:57
• Apply Kolmogorov's three series theorem or the martingale convergence theorem to prove convergence of the mean zero version $\sum_{k=1}^n (X_k-\xi)/k$, and then conclude. – user940 Nov 3 '13 at 15:03