Solution: Problem1-57
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Solution: Problem1-57
siromasiroma 1222339519|%e %b %Y, %H:%M %Z|agohover

Geometric progression sum formula derivation:

(1)
S_n=\sum_{n=0}^\infty {q^{n}a_0}=a_0+qa_0+q^2a_0+...+q^na_0
(2)
qS_n=q\sum_{n=0}^\infty {q^na_0}=qa_0+q^2a_0+q^3a_0+...+q^{n+1}a_0
(3)
(q-1)S_n=q^{n+1}a_0-a_0
(4)
S_n=a_0\frac{q^{n+1}-1}{q-1}

If q<1 then

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

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)
\lim_{n \to \infty } S_n = \frac{1}{1-e^{-\alpha}}

On the other hand,

(7)
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)

Because \alpha>0

(8)
I = \frac{1}{\alpha}

To find required \alpha we have

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

or

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

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

(11)
\alpha=0

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

(12)
I \to S \text{ when } \alpha \to 0^+
last edited on 1222342030|%e %b %Y, %H:%M %Z|agohover by siroma + show more
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