I saw following exercise from Durrett's probability theory book and I managed to solve the 1st part, but couldn't get the 2nd part.

Let $X_1, X_2, \dots$ be i.i.d samples with mean $0$, and finite non-zero variance. Denote $S_n = X_1 + X_2+\dots + X_n$.

  1. Use central limit theorem and Kolmogorov $0-1$ law to show $\limsup S_n/\sqrt{n} = \infty$ a.s.
  2. Use contradiction to show $S_n/\sqrt{n}$ does not converge in probability. (Hint: consider $n=m!$)

For 1st part: (Please point out errors if you see any)

W.L.O.G, we may assume $Var(X_1)=1$. Then by C.L.T, we have $S_n/\sqrt{n} \sim Z=N(0,1)$. Then the event $$\left\{\limsup S_n/\sqrt{n} = \infty\right\} = \left\{\bigcap _{m=1}^{\infty} \bigcup_{n=m}^{\infty} S_n/\sqrt{n} = \infty\right\}$$ is in the tail $\sigma$-algebra formed by $X_1,X_2,\dots$. By $0-1$ law, we have $P(\limsup S_n/\sqrt{n} = \infty)$ is either $0$ or $1$. Now, the event $\{\limsup S_n/\sqrt{n} = \infty\}$ is equivalent to: for any $N$, we have $\{\limsup S_n/\sqrt{n} > N\}$. Then \begin{align} P\left(\limsup S_n/\sqrt{n} > N\right)&=\lim_{m\rightarrow \infty} P\left(\bigcup_{n=m}^\infty S_n/\sqrt{n} > N\right) \\[0.2cm] & \ge \lim_{m\to \infty} P\left( S_m/\sqrt{m} > N\right) \\[0.3cm] & = 1-\Phi(N) >0 \end{align}

For 2nd part: Suppose it converges, then it must converge to normal $Z$. Then I am not sure how to proceed. I think it may somehow connect with part $1$. The hint also reminds me of Stirling's approximation, not sure.


  • 1
    $\begingroup$ @Jimmy R. Many thanks for your editing. $\endgroup$
    – Lei Hao
    Nov 25, 2016 at 11:46

1 Answer 1


For the first, you need more details at the end: we have to use the fact that for each $N$, the event $\left\{\limsup S_n/\sqrt{n} > N\right\}$ is a tail event and by the previous work, its probability is $1$. Therefore, the intersection has a probability equal to $1$.

For the second part, we can show that the sequence $\left(S_{m!} / \sqrt{m!}\right)_{m\geqslant 1}$ does not converge in probability. Otherwise this sequence converges to some $Y$ in probability, we would have $S_{(m+1)!} / \sqrt{(m+1)!}-S_{m!} / \sqrt{m!}\to 0$ in probability. Define $$U_m:=\left(\frac 1{\sqrt{(m+1)!}} -\frac 1{\sqrt{m!}} \right) S_{m!} $$ $$V_m :=\frac 1{\sqrt{(m+1)!}}\left(S_{(m+1)!}-S_{m!}\right).$$ Then the sequences $\left(U_m\right)_{m\geqslant 1}$ and $\left(V_m\right)_{m\geqslant 1}$ are independent. Since for any positive $\epsilon$, $$\mathbb P\left(\left|U_m+V_m\right | \gt \epsilon\right)\geqslant \mathbb P\left(\left\{\left|U_m\right | \gt 2\epsilon\right\}\cap \left\{\left|V_m\right|\leqslant\epsilon \right\}\right),$$ we get that $U_m\to 0$ in probability.

  • $\begingroup$ Many thanks for your answer and pointing out my errors. One more question regarding the 2nd part, what is the role of the factorial? It seems the argument can be applied to n and n+1 directly. Thanks. $\endgroup$
    – Lei Hao
    Nov 26, 2016 at 8:14
  • $\begingroup$ @LeiHao Maybe there is something I have not seen. It works fine with $n$ and $2n$ for example (not sure with $n$ and $n+1$). $\endgroup$ Nov 26, 2016 at 11:42

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