limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n)=
∫2 to 5 4√xⅆx
To find this limit, we can transform the sum into a definite integral using the limit definition of a Riemann sum. Specifically, we can write:
lim n→∞ ∑ k=1n (2 + 3k/n^(1/2)) / (4⋅3^(n/2))
= lim n→∞ [1/n] ∑ k=1n (2 + 3k/n^(1/2)) / (4⋅3^(n/2))
= lim n→∞ [1/n] ∑ k=1n [(2/n^(1/2)) + (3/n)^(1/2)]
= ∫₀¹ [(2/x^(1/2)) + (3x)^(1/2)] / (4⋅3^x) dx, where we have used the substitution x = k/n and taken the limit as n goes to infinity.
We can then evaluate this integral using standard techniques, such as u-substitution. Letting u = (3x)^(1/2), we have du/dx = 3/2 u^(−1/2), so dx = 2/3 u^(1/2) du. Substituting u = (3x)^(1/2) and dx = 2/3 u^(1/2) du, we get:
lim n→∞ ∑ k=1n (2 + 3k/n^(1/2)) / (4⋅3^(n/2))
= ∫₀¹ [(2/x^(1/2)) + (3x)^(1/2)] / (4⋅3^x) dx
= ∫₀^(√3) [(2/u^2) + u] / (4⋅3^(u^2/3)) du
= [1/4] ∫₀^(√3) [(u^−2) + 3^(u^2/3-2)] du
= [1/4] [(∞ − 1) + (3^(1/3) − 1) + ∫₀^(√3) 3^(u^2/3-2) du]
The first two terms in the brackets are simply limits of the integrand as u approaches 0 and √3, respectively. The last term can be evaluated using the substitution v = u^(2/3), so that dv/dx = (2/3) u^−1/3 and dx = (3/2) v^1/3 dv. Substituting v = u^(2/3) and dx = (3/2) v^1/3 dv, we get:
lim n→∞ ∑ k=1n (2 + 3k/n^(1/2)) / (4⋅3^(n/2))
= [1/4] [(∞ − 1) + (3^(1/3) − 1) + ∫₀¹ 3^(v-2/3) dv]
= [1/4] [(∞ − 1) + (3^(1/3) − 1) + (3^(1/3)/ln(3) − 1/ln(3))]
= (3^(1/3) + 3^(−2/3) − 2)/(4ln(3))
Therefore, the limit of the sum as n approaches infinity evaluates to:
lim n→∞ ∑ k=1n (2 + 3k/n^(1/2)) / (4⋅3^(n/2)) = (3^(1/3) + 3^(−2/3) − 2)/(4ln(3)).
Note that this limit exists and is finite, as both the numerator and denominator approach finite values.
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