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Let consider $0<a_1<a_2<...<a_n$. What are the following limits $$ \begin{array}{l} A = \mathop {\lim }\limits_{r \to \infty } \frac{{\sum\limits_{i = 1}^n {{e^{ - {a_i}r}}} }}{{\sum\limits_{i = 1}^n {a_i}{e^{ - {a_i}r}}} }\\ B = \mathop {\lim }\limits_{r \to \infty } \frac{{\sum\limits_{i = 1}^n {a_i^2{e^{ - {a_i}r}}} }}{{\sum\limits_{i = 1}^n {a_i}{e^{ - {a_i}r}}} } \end{array} $$

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What is $a_i$?${}$ – Eckhard Jul 9 '13 at 8:43
$a_i$ is just a finite constant – widapol Jul 9 '13 at 18:05
up vote 2 down vote accepted

Hint: Since $0 < a_1 < a_2 < \cdots < a_n$ and $e^{-rx}$ is a monotone decreasing function in $x$ (where $r$ is some fixed positive constant), we know that the expression: $$ e^{-a_1r} $$ dominates all other expressions of the form $e^{-a_jr}$ (where $j \in \{2,3,...,n\}$) as $r\to\infty$. So the limits $A$ and $B$ will intuitively be the ratio of the coefficients of these expressions. That is, you will find that: $$ A=\dfrac{1}{a_1} \qquad\text{and}\qquad B=\dfrac{a_1^2}{a_1}=a_1 $$

To see this, multiply the numerator and denominator by $e^{a_1r}$ and exploit the fact that for any $j \in \{2,3,...,n\}$ and for any $k \in \Bbb{R}$: $$ \lim_{r\to\infty} \dfrac{k}{e^{(a_j-a_1)r}}=0 $$

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$$A=\frac{1}{a_1}$$ $$B=a_1$$

$$ $$

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+1 for the very compact hint :) – Babak S. Jul 9 '13 at 9:48
Proof? Hint? Sketch? ${}{}{}{}$ – Pedro Tamaroff Jul 11 '13 at 5:26

Hint for $A$: $$\lim\limits_{r\rightarrow+\infty}\frac{e^{-a_1r}\left(1+\sum\limits_{j=2}^ne^{(a_1-a_j)r}\right)}{e^{-a_1r}\left(a_1+\sum\limits_{j=2}^n a_je^{(a_1-a_j)r}\right)}=\frac{1}{a_1}$$ $$\lim\limits_{r\rightarrow-\infty}\frac{e^{-a_nr}\left(1+\sum\limits_{j=1}^{n-1}e^{(a_n-a_j)r}\right)}{e^{-a_nr}\left(a_n+\sum\limits_{j=1}^{n-1} a_je^{(a_n-a_j)r}\right)}=\frac{1}{a_n}$$

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This is a solution, not a hint. – Pedro Tamaroff Jul 11 '13 at 5:27

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