I got stuck trying to solve this limit. <munder> <mo movablelimits="true" form="prefix">lim

Esmeralda Lane 2022-07-07 Answered
I got stuck trying to solve this limit.
lim n 1 + 1 2 + 1 4 + . . . . + 1 2 n 1 + 1 3 + 1 9 + . . . + 1 3 n
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Answers (1)

behk0
Answered 2022-07-08 Author has 14 answers
We see, that ( 1 + 1 2 + 1 4 + 1 8 + + 1 2 n ) and ( 1 + 1 3 + 1 9 + . . . + 1 3 n ) are the geometric series.
Therefore,
lim n ( 1 + 1 2 + 1 4 + 1 8 + + 1 2 n ) = n = 0 ( 1 2 ) n = 1 1 1 2 = 2 ,
lim n ( 1 + 1 3 + 1 9 + . . . + 1 3 n ) = n = 0 ( 1 3 ) n = 1 1 1 3 = 3 2
We will have
lim n ( 1 + 1 2 + 1 4 + 1 8 + + 1 2 n ) ( 1 + 1 3 + 1 9 + . . . + 1 3 n ) = 2 3 2 = 4 3 .

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