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Evaluate the limit as n tends to infinity of the sum 1/(2n+1) + 1/(2n+2) +... + 1/(3n).

1. logₑ(1/3)
2. logₑ(2/3)
3. logₑ(3/2)
4. logₑ(4/3)

Correct answer: logₑ(3/2)

Solution

Write the sum as (1/n) * sum over r=1..n of 1/(2+r/n), a Riemann sum for the integral from 0 to 1 of dx/(2+x) = ln(3/2). So the limit is ln(3/2).

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