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Consider $X_m$ which are independent, identically distributed random variables which have moments exactly up to order $2$ and no higher. This can be done in numerous ways. One is the following: let $a_n$ be a sequence of positive numbers that goes to zero and let $b_n$ be a sequence of positive numbers such that $\sum_{n=1}^\infty \frac{b_n}{a_n}$ converges. (For example, $a_n=1/n,b_n=2^{-n}$ does the job.) Let $f_n(x)=(2+a_n) 1_{[1,\infty)}(x) |x|^{-3-a_n}$. Then let $f(x)=\sum_{n=1}^\infty \frac{b_n}{\sum_{k=1}^\infty b_k} f_n(x)$. Then $f$ is a pdf of a r.v. which has no higher moments than the second (since each $f_n$ has exactly moments of order strictly less than $2+a_n$). The assumption about $b_n$ ensures that it actually does have a finite second moment. (In effect the trick here is $\bigcap_{n=1}^\infty [1,2+a_n)=[1,2]$.)

Despite this somewhat pathological moment property, the classical central limit theorem still tells us that $\frac{\overline{X}_m-\mu}{\sigma/\sqrt{m}}$ converges in distribution to a $N(0,1)$ random variable. However, the most common quantitative estimates for the convergence rate, e.g. the Berry-Esseen theorem, require the existence of a moment higher than order $2$. What can be said about the convergence rate in cases like the one in the previous paragraph?

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The paper

Friedman, N.; Katz, Melvin; Koopmans, L. H. Convergence rates for the central limit theorem. Proc. Nat. Acad. Sci. U.S.A. 56 1966 1062–1065.,

available here, seems to give something in this direction. The authors prove that for an i.i.d. sequence of centered random variables with unit variance and such that $\mathbb E\left[X_1^2\log\left(1+\left|X_1\right|\right)\right]< +\infty$, the series $$\sum_{n=1}^{+\infty } \frac 1n\left|\mathbb P\left\{ \frac{S_n } { \sqrt n }\leqslant x\right\} - \Phi(x) \right| $$ is convergent for any $x\in\mathbb R\setminus \{0\}$, where $S_n=\sum_{i=1}^nX_i$ and $\Phi$ denotes the cumulative distribution function of a standard normal distribution.

However, we still have a little bit more than a finite second moment.

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    $\begingroup$ Unless I am mistaken, for an i.i.d. sequence of centered random variables with unit variance, the authors show that the convergence of the series in your post is equivalent to the convergence of$$\sum_n\frac1n\left(E(X_1^2;|X_1|\geqslant\sqrt{n})+E(X_1;|X_1|\leqslant\sqrt{n})^2\right)$$ which seems to be equivalent (but I fail to understand why the authors do not mention this) to the finiteness of $$\sum_n\frac1nE(X_1^2;|X_1|\geqslant\sqrt{n})$$ or, even more simply, of $$E(X_1^2\log^+|X_1|).$$ (That the case $x=0$ is different, as Rosen's theorem shows, is notable.) $\endgroup$ Commented Jul 31, 2016 at 9:56
  • $\begingroup$ @Did You are right. I should have looked at the paper more carefully. It is an element of answer, but of course not a complete answer to the question. $\endgroup$ Commented Jul 31, 2016 at 12:21
  • $\begingroup$ We might be closer to a definitive answer than your comment makes believe since every square integrable random variable $Y$ is such that $E(u(Y)Y^2)$ is finite, for some well chosen function $u$ going to infinity at infinity. $\endgroup$ Commented Jul 31, 2016 at 14:56
  • $\begingroup$ Thanks, that is helpful and relevant. It also seems to suggest that we might expect very little here. In particular, since the result excludes $x=0$, I would expect that it is not uniform in $x$. Additionally, the convergence of that sum is a rather mild property, for instance it is enough to have the difference being $O(\log(n)^{-1-\delta})$... $\endgroup$ Commented Jul 31, 2016 at 14:57
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    $\begingroup$ @Ian In full generality, pick $y_0=0$ and $(y_n)$ nondecreasing such that $E(Y^2;|Y|\geqslant y_n)\leqslant1/n^3$ for every $n\geqslant1$ and define $u(y)=n$ if $y_n\leqslant|y|<y_{n+1}$, then $u\to\infty$ at infinity and $E(Y^2u(Y))\leqslant\sum\limits_nE(Y^2u(Y);y_n\leqslant|Y|< y_{n+1})\leqslant\sum\limits_nnE(Y^2;y_n\leqslant|Y|< y_{n+1})\leqslant\sum\limits_nnE(Y^2;|Y|\geqslant y_{n})$ converges. In your $(a_n,b_n)$-case, I did not check whether this construction yields anything specific. $\endgroup$ Commented Jul 31, 2016 at 15:05

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