What is <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD"

Janet Forbes 2022-07-12 Answered
What is lim n n ( k = 1 n 1 n 2 + k 1 ) ?
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Answers (2)

Ashley Parks
Answered 2022-07-13 Author has 11 answers
We can rewrite the limit as
lim n k = 1 n n n 2 + k n = lim n k = 1 n n n 2 + k n 2 + k = lim n k = 1 n 1 1 + k n 2 1 + k n 2
Since the numerator approaches zero we will rationalize and continue to simplify
lim n k = 1 n k n 2 1 + k n 2 + 1 + k n 2 = lim n k = 1 n k n 1 + k n 2 + 1 + k n 2 1 n
We can sandwich this complicated expression with two other limits
lim n 1 2 k = 1 n k n 1 n < lim n k = 1 n k n 1 + k n 2 + 1 + k n 2 1 n < lim n 1 1 + 1 n + 1 + 1 n k = 1 n k n 1 n
Both of which approach the Riemann sum
1 2 0 1 x d x = 1 4
Thus the limit is 1 4 by squeeze theorem.
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malalawak44
Answered 2022-07-14 Author has 4 answers
We have
n ( k = 1 n 1 n 2 + k 1 ) = n k = 1 n ( 1 n 2 + k 1 n ) = n k = 1 n n n 2 + k n n 2 + k = k = 1 n k n 2 + k ( n + n 2 + k )
and, for 1 k n
1 ( n + 1 ) ( 2 n + 1 ) 1 n 2 + k ( n + n 2 + k ) 1 n ( 2 n )
while k = 1 n k = n ( n + 1 ) / 2. Thus (momentarily dropping the minus sign in the numerator k) we have
n 2 ( n + 1 ) ( 2 n + 1 ) k = 1 n k n 2 + k ( n + n 2 + k ) ( n + 1 ) 4 n
so by the squeeze theorem (and putting the minus sign back in), we have
lim n n ( k = 1 n 1 n 2 + k 1 ) = 1 4
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