A geometric series is the sum of the terms of a geometric sequence of the form [tex]a, ar, ar^{2}, ar^{3},... [/tex],Â
where r is the common ratio, and a≠0 is the first term.
That is, the series is  [tex]a+ar+ar^{2}+ar^{3},... [/tex]
In sigma notation, the series is written as:
∞
∑  [tex]a r^{k} [/tex]Â
k=0
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The geometric series of the form
∞
∑  [tex]a r^{k} [/tex], converges to [tex] \frac{a}{1-r} [/tex] if |r|<1
k=0
and diverges otherwise.
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in our problem, a , the first term is equal to 940, and the common ratio is |1/5|<1,Â
thus the series converges to:
[tex] \frac{a}{1-r}=\frac{940}{1-1/5}=\frac{940}{4/5}=752[/tex]
Answer:Â
∞       Â
∑  [tex]940 (1/5)^{k} [/tex]=752  ( the upper limit of the population is 752)
k=0
