Theorem 5 (modified from Gnedenko (1943), pp. 438-439)

Let R and S be real numbers such that

R = e^-S,

and let $n_1 < n_2 < n_3 < \ldots$ be any subsequence of the positive integers. For x = x₀ and for any finite real constants a_nj > 0 and b_nj, $j = 1,2,3,\ldots$,

 

$$\lim_{j \to \infty} F^{n_j} (a_{n_j} x_0 + b_{n_j} ) = R$$ (29)

if and only if

 

$$\lim_{j\to\infty} n_j \left[ 1 - F (a_{n_j} x_0 + b_n) \right] = S.$$ (30)

Proof: By Taylor's Theorem [Landau (1951), p. 120, theorem 178],

$$\log y = - (1-y) \left\{ 1 - \frac{1-y}{2q^2} \right\}, \qquad y < q < 1.$$

Then,

 

$$\begin{array} {r@{}l} \log F^{n_j} & (a_{n_j} x_0 + b_{n_j})... ...frac{1 - F (a_{n_j} x_0 + b_{n_j} )}{2q^2} \right\},\end{array}$$ (31)

where F(a_nj x₀ + b_nj ) < q < 1.

(a) Suppose

$$\lim_{j \to \infty} F^{n_j} (a_{n_j} x_0 + b_{n_j}) = R.$$

If $R \neq 0$, then by Eq. (29),

$$\lim_{j\to\infty} F(a_{n_j} x_0 + b_{n_j} ) = 1.$$

Hence, taking the limit in Eq. (31) gives

 

$$\lim_{j\to\infty} \log F^{n_j} ( a_{n_j} x_0 + b_{n_j}) = - \lim_{j\to\infty} n_j \left[ 1-F (a_{n_j} x_0 + b_{n_j} ) \right]$$ (32)

or $- \log R = S$. If R = 0 and $\lim_{j \to \infty} F(a_{n_j} x_0 + b_{n_j} ) = 1$,the above argument still yields Eq. (32).

If R = 0 and $\overline{\lim} _{j\to\infty} F (a_{n_j} x_0 + b_{n_j}) < 1$, then immediately

$$\begin{array}[t] {c} \underline{\lim} \\ \scriptstyle j \to... ...1 - F (a_{n_j} x_0 + b_{n_j} ) \right] = -\infty = -\log R = S.$$

Hence, in all cases,

 

$$\lim_{j \to \infty} n_j \left[ 1 - F (a_{n_j} x_0 + b_{n_j} ) \right] = -\log R = S.$$ (33)

(b) Now, suppose that

$$\lim_{j\to\infty} n_j \left[ 1-F (a_{n_j} x_0 + b_{n_j}) \right] = S.$$

If $S < \infty$, then $\lim_{j \to \infty} F(a_{n_j} x_0 + b_{n_j} ) = 1$, and by Eq. (31) as $n \to \infty$,

$$\lim_{j\to\infty} \log F^{n_j} (a_{n_j} x_0 + b_{n_j}) = - \... ...to\infty} n_j \left[ 1 - F (a_{n_j} x_0 + b_{n_j}) \right] = -S$$

or

 

$$\lim_{j\to\infty} F^{n_j} (a_{n_j} x_0 + b_{n_j}) = e^{-S}.$$ (34)

If $S = \infty$ and $\lim_{j \to \infty} F(a_{n_j} x_0 + b_{n_j} ) = 1$, the reasoning leading to Eq. (34) still holds and Eq. (34) is true. If $S = \infty$ and $\lim_{j\to\infty} F (a_{n_j} x_0 + b_{n_j} ) < 1$, then

$$\begin{array}[t] {c} \overline{\lim} \\ \scriptstyle j\to\infty\end{array} F^{n_j} (a_{n_j} x_0 + b_{n_j} ) = 0 = e^{-S}.$$

Hence, in every case,

$$\lim_{j\to\infty} F^{n_j} (a_{n_j} x_0 + b_{n_j} ) = e^{-S} = R.$$