How can I solve <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeX

Jerry Villegas 2022-05-30 Answered
How can I solve
lim n 1 n 2 k = 0 n 1 ln ( 1 + ( n + k ) n 2 + k 2 n 3 )
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Answers (1)

iberistazi
Answered 2022-05-31 Author has 10 answers
lim n 1 n 2 k = 0 n 1 log ( 1 + ( n + k ) n 2 + k 2 n 3 ) = lim n 1 n 2 k = 0 n 1 log ( 1 + 1 n ( 1 + k n ) 1 + k 2 n 2 ) .
Now, remembering that x log ( 1 + x ) for every x > 0 and that for every ε > 0 there is a δ > 0 such that for every x ( 0 , δ ) we have log ( 1 + x ) ( 1 ε ) x
The first inequality implies
lim n 1 n 2 k = 0 n 1 log ( 1 + 1 n ( 1 + k n ) 1 + k 2 n 2 ) lim n 1 n 2 k = 0 n 1 1 n ( 1 + k n ) 1 + k 2 n 2 =
= lim n 1 n k = 0 n 1 ( 1 + k n ) 1 + k 2 n 2 = 0 1 1 ( 1 + x ) 1 + x 2 d x .
On the other hand, since (for a fixed ε) for n big enough we have that for every k it holds that 1 n ( 1 + k n ) 1 + k 2 n 2 2 2 n δ
lim n 1 n 2 k = 0 n 1 log ( 1 + 1 n ( 1 + k n ) 1 + k 2 n 2 ) 1 1 ε 0 1 1 ( 1 + x ) 1 + x 2 d x .
Since ε is arbitrary, we get
lim n 1 n 2 k = 0 n 1 log ( 1 + 1 n ( 1 + k n ) 1 + k 2 n 2 ) = 0 1 1 ( 1 + x ) 1 + x 2 d x .
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