View source for Lindeberg-Feller theorem

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A theorem that establishes necessary and sufficient conditions for the asymptotic normality of the distribution function of sums of independent random variables that have finite variances. Let  $  X _ {1} , X _ {2} \dots $
be a sequence of independent random variables with means  $  a _ {1} , a _ {2} \dots $
and finite variances  $  \sigma _ {1}  ^ {2} , \sigma _ {2}  ^ {2} \dots $
not all of which are zero. Let

$$ 
B _ {n}  ^ {2}  =  \sum _ { j= } 1 ^ { n }  \sigma _ {j}  ^ {2} ,\ \ 
V _ {j} ( x)  =  {\mathsf P} \{ x _ {j} < x \} .
$$

In order that

$$ 
B _ {n}  ^ {-} 2  \max _ {1 \leq  j \leq  n } \ 
\sigma _ {j}  ^ {2}  \rightarrow  0
$$

and

$$ 
{\mathsf P} \left \{ B _ {n}  ^ {-} 1 \sum _ { j= } 1 ^ { n } 
( X _ {i} - a _ {j} ) < x \right \}  \rightarrow \ 

\frac{1}{\sqrt {2 \pi }}
 \int\limits _ {- \infty } ^ { x } 
e ^ {- t  ^ {2} /2 }  d t
$$

for any  $  x $
as  $  n \rightarrow \infty $, 
it is necessary and sufficient that the following condition (the Lindeberg condition) is satisfied:

$$ 
B _ {n}  ^ {-} 2 \sum _ { j= } 1 ^ { n } 
\int\limits _ {| x - a _ {j} | \geq  \epsilon B _ {n} }
( x - a _ {j} )  ^ {2}  d V _ {j} ( x)  \rightarrow  0
$$

as  $  n \rightarrow \infty $
for any  $  \epsilon > 0 $. 
Sufficiency was proved by J.W. Lindeberg [[#References|[1]]] and necessity by W. Feller [[#References|[2]]].

====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  J.W. Lindeberg,   "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung"  ''Math. Z.'' , '''15'''  (1922)  pp. 211–225</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  W. Feller,   "Ueber den zentralen Grenzwertsatz der Wahrscheinlichkeitsrechnung"  ''Math. Z.'' , '''40'''  (1935)  pp. 521–559</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  M. Loève,   "Probability theory" , Springer  (1977)</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  V.V. Petrov,   "Sums of independent random variables" , Springer  (1975)  (Translated from Russian)</TD></TR></table>