A Technical Lemma

The following lemma, which is an elementary analysis result, will be needed for calculating the limit of E[X_n,a] in the next section.

Lemma 1   Suppose $\alpha > \beta\geq 0$. Let (L_n) and (M_n) be sequences of positive integers which satisfy

$$\frac{L_n}{n} \rightarrow L >0$$

and

$$\frac{M_n}{n} \rightarrow M >0.$$

Then the sum

$$\sum_{k=L_n}^{M_n} \frac{1}{\alpha k + \beta} = \frac{1}{\alpha} \ln (M_n) - \frac{1}{\alpha} \ln (L_n) + A_n,$$

where

$$\vert A_n\vert \leq \frac{1}{\alpha L_n}.$$

In particular,

$$\lim_{n\rightarrow \infty} \sum_{k=L_n}^{M_n} \frac{1}{\alpha k + \beta} = \frac{1}{\alpha} \ln\left(\frac{M}{L}\right).$$

 

Proof. First, observe that for any integers 0 < x < y, the sum

$$\sum_{k = x + 1}^y\frac{1}{k} = \ln (y/x) + \epsilon_1 (x, y),$$ (21)

where

$$\frac{1}{2y} -\frac{1}{2x} \le \epsilon_1 (x, y) \le 0.$$ (22)

(This can be seen by comparing the sum with the integral $\int_x^y \frac{1}{t} dt$.)

Next, since $\alpha > \beta\geq 0$, note that

$$\sum_{k = x + 1}^{y+1}\frac{1}{\alpha k} < \sum_{k = x}^y\frac{1}{\alpha k+\beta} \le \sum_{k = x}^y\frac{1}{\alpha k}.$$ (23)

Thus, using (21),

$$\sum_{k = x}^y\frac{1}{\alpha k+\beta} ={1 \over \alpha}\ln(y/x)+\epsilon_2(x,y),$$ (24)

where

$$\frac{1}{\alpha}\left(\frac{1}{y+1}+\epsilon_1(x,y)\right) < ... ...)\le \frac{1}{\alpha}\left(\frac{1}{x}+\epsilon_1(x,y)\right).$$ (25)

Combining this with (22), the error can be bounded by

$$\vert \epsilon_2 (x, y) \vert \le\frac{ 1}{\alpha x}.$$ (26)

Setting $A_n = \epsilon_2 (L_n, M_n)$,

$$\sum_{k=L_n}^{M_n} \frac{1}{\alpha k + \beta} = \frac{1}{\alpha} \ln (M_n) - \frac{1}{\alpha} \ln (L_n) + A_n.$$ (27)

Finally, note that since $L_n/n \rightarrow L>0$, the sequence $1/(\alpha L_n) \rightarrow 0$. Thus $A_n \rightarrow 0$, and

$$\lim_{n\rightarrow \infty} \sum_{k=L_n}^{M_n} \frac{1}{\alpha k + \beta} = \frac{1}{\alpha} \ln\left(\frac{M}{L}\right).$$