Solution: Problem1-57

siroma 25 Sep 2008 10:45

Geometric progression sum formula derivation:

(1)\begin{align} S_n=\sum_{n=0}^\infty {q^{n}a_0}=a_0+qa_0+q^2a_0+...+q^na_0 \end{align}

(2)
\begin{align} qS_n=q\sum_{n=0}^\infty {q^na_0}=qa_0+q^2a_0+q^3a_0+...+q^{n+1}a_0 \end{align}

(3)
\begin{equation} (q-1)S_n=q^{n+1}a_0-a_0 \end{equation}

(4)
\begin{align} S_n=a_0\frac{q^{n+1}-1}{q-1} \end{align}

If q<1 then

(5)\begin{align} \lim_{n \to \infty } S_n = a_0\frac{1}{1-q} \end{align}

In the problem $a_0=1$ and $q=e^{-\alpha}$ therefore we require that $e^{-\alpha} < 1$; thus when $\alpha>0$ we have closed sum

(6)\begin{align} \lim_{n \to \infty } S_n = \frac{1}{1-e^{-\alpha}} \end{align}

On the other hand,

(7)\begin{align} I=\int_{0}^{\infty}{e^{-\alpha n}}dn = -\frac{e^{-\alpha n}}{\alpha} |_0^\infty = -\frac{1}{\alpha}(\lim_{n \to \infty} e^{-\alpha n} - 1) \end{align}

Because $\alpha>0$

(8)\begin{align} I = \frac{1}{\alpha} \end{align}

To find required $\alpha$ we have

(9)\begin{align} \frac{1}{\alpha}=\frac{1}{1-e^{-\alpha}} \end{align}

or

(10)\begin{align} e^{-\alpha}=1-\alpha \end{align}

Equation (10) has only one solution(this can be shown comparing derivatives of rhs and lhs functions):

(11)\begin{align} \alpha=0 \end{align}

To satisfy requirement of $\alpha > 0$ we give result in limit form:

(12)\begin{align} I \to S \text{ when } \alpha \to 0^+ \end{align}