n → ∞ lim [ ln ( n n 2 4 ) + ln ( n n 2 1 6 ) + ln ( n n 2 3 6 ) + ⋯ + ln ( n n 2 4 n 2 ) ] = ?
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In the last step, i think you should limit it first because ln(x) tends to -infinite as x tends to 0
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lim x → 0 x ln x = lim x → 0 x 1 ln x = − x 2 1 x 1 = 0 by L'Hopital rule.
I think you can use Stirling's approximation you get
L = n → ∞ lim k = 1 ∑ n ln ( n n 2 4 k 2 ) = n → ∞ lim ln ⎝ ⎜ ⎜ ⎜ ⎛ n n 2 n k = 1 ∏ n 4 k 2 ⎠ ⎟ ⎟ ⎟ ⎞ = n → ∞ lim ln ( n 2 4 n k = 1 ∏ n k 2 ) = n → ∞ lim ln ( n 2 4 n n ! n ! ) = n → ∞ lim ln ( n 2 4 e 2 n 2 n 2 π n ) = ln ( e 2 4 ⋅ 1 ⋅ 1 ) = 2 ln ( 2 ) − 2
edit: If you think, it is too inexact because we use a approximation: No, because n → ∞ :
More exactly you have: n ! = 2 π n ( e n ) n e μ ( n ) with 0 < μ ( x ) < 1 / ( 1 2 x ) , x > 0 ⇒ n → ∞ lim μ ( n ) = 0 , and so:
L 2 = n → ∞ lim ln ( n n ! n ! ) = n → ∞ lim ln ( e 2 n 2 n 2 π n ) + n → ∞ lim ln ( e 2 μ / n ) = n → ∞ lim ln ( e 2 n 2 n 2 π n )
I think of Riemann sum and this one too but unable to use Stirling approximation . Thanks !!
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You're welcome.
I have added something for better understanding, btw.
Rearranging terms and using the log identity ln(a)+ln(b)+...ln(z) = ln(ab...z) gives us n → ∞ lim ( n 2 ln ( n n 2 n n ! ) ) Using Stirling's approximation for n! ( 2 π n n + 2 1 e − n ≤ n ! ≤ e n n + 2 1 e − n ), we get n → ∞ lim 2 ln ( e 2 n k n ) where 2 π ≤ k ≤ e . This gives us the desired result 2 ln 2 − 2
Amateur explanation:
This can be rewritten as a definite integral with bounds 0 to 2 as shown:
lim n → ∞ ∑ i = 1 n l n ( ( n 2 i ) ( n 2 ) )
= lim n → ∞ ∑ i = 1 n ( n 2 ) l n ( n 2 i )
= lim n → ∞ ( n 2 ) ∑ i = 1 n l n ( n 2 i )
= ∫ 0 2 l n ( x ) d x
= 2 l n ( 2 ) − 2
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Relevant wiki: Riemann Sums
L = n → ∞ lim [ ln ( n n 2 4 ) + ln ( n n 2 1 6 ) + ln ( n n 2 3 6 ) + ⋯ + ln ( n n 2 4 n 2 ) ] = n → ∞ lim k = 1 ∑ n ln ( n n 2 4 k 2 ) = n → ∞ lim k = 1 ∑ n n 1 ln ( n 2 4 k 2 ) = ∫ 0 1 ln ( 4 x 2 ) d x = ∫ 0 1 ( 2 ln 2 + 2 ln x ) d x = 2 x ln 2 + 2 x ln x − 2 x ∣ ∣ ∣ ∣ 0 1 = 2 ln 2 + 2 ln 1 − 2 x − 0 − 2 x → 0 lim x ln x − 0 = 2 ln 2 − 2 By Riemann sums: n → ∞ lim k = a ∑ b n 1 f ( n k ) = k → n lim ∫ k a k b f ( x ) d x See note: x → 0 lim x ln x = 0
Note:
L = x → 0 lim x ln x = x → 0 lim x 1 ln x = x → 0 lim − x 2 1 x 1 = x → 0 lim − x = 0 A ∞ / ∞ cases, L’H o ˆ pital’s rule applies. Differentiate up and down w.r.t x .