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Leklerk
@Leklerk
October 2021
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[tex] \lim_{n \to \infty} \frac{1+ \frac{1}{3}+ \frac{1}{ 3^{2} } +...+ \frac{1}{3^{n} } }{1+ \frac{1}{5} + \frac{1}{ 5^{2} }+...+ \frac{1}{ 5^{n} } }[/tex]
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В числителе и знаменателе имеем сумму убывающей геометрической прогрессии.
S=b/(1-q)
1/(1-1/3)=3/2
1/(1-1/5)=5/4
lim=(3/2)/(5/4)=6/5
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Answers & Comments
Verified answer
В числителе и знаменателе имеем сумму убывающей геометрической прогрессии.S=b/(1-q)
1/(1-1/3)=3/2
1/(1-1/5)=5/4
lim=(3/2)/(5/4)=6/5