Let $X_m$ be distributed as follows: $$\mathbb P(X_m=\pm m)=\frac{1}{2m^\beta}, \mathbb P(X_m=0)=\frac{m^\beta-1}{m^\beta}.$$ Let be $S_n=\sum_{m=1}^{n}X_m$, then
I. If $\beta>1$, then $\exists S_\infty$ s.t. $S_n\to S_\infty$ almost surely;
II. If $\beta<1$, then $$\frac{S_n}{c*n^{(3-\beta)/2}}\implies \mathscr N(0,1);$$
III. If $\beta=1$, then $$\mathbb E\left(e^{itS_n/n}\right)\to exp\left(-\int_{0}^{1}\frac{1-cos(ts)}{s}ds\right).$$
I started with calculating the characteristic function of $S_n$: $$\mathbb E\left(e^{itS_n}\right)=\prod_{m=1}^{n}\mathbb E\left(e^{itX_m}\right)=\prod_{m=1}^{n}\frac{1}{m^\beta}\left(\frac{e^{itm}+e^{-itm}}{2}+m^\beta-1\right)=\prod_{m=1}^{n}\frac{cos(tm)+m^\beta-1}{m^\beta}$$
My thought process is that if the limit of the characteristic functions is calculated, then it should converge to:
a constant if $\beta>1$, and it would imply almost sure convergence;
characteristic function of a normal distribution if $\beta<1$
the given integral if $\beta=1$.
My problem is that I cannot determine the limit of that product. Could you help me with that?