Definite Integral As Limit Of Sum
By treating the expression as a Riemann sum, evaluate the limit as n tends to \infty of the sum of n divided by the quantity n-squared plus k-squared for k from one to n.
Select the correct option:
Solution
π/4
A limit of a sum indexed from 1 to n with a 1/n scaling factor is the signature of the Riemann-sum definition, which lets us replace a hard discrete sum by a continuous definite integral. The strategy is to factor out 1/n as the sample spacing and identify k/n as the sample point, so we rewrite \sum_{k=1}^{n}\frac{n}{n^2+k^2} = \sum_{k=1}^{n}\frac{1}{n}\cdot\frac{1}{1 + (k/n)^2}. As n \to \infty, setting x = k/n and dx = 1/n, the sum converges to \int_0^1 \frac{dx}{1 + x^2} = [\arctan x]_0^1 = \arctan 1 - \arctan 0 = \pi/4. Option \pi/2 would arise from integrating over [0, \infty) instead of the unit interval. Option 1 ignores the arctangent structure of the integrand. Option \ln 2 corresponds to a different integrand such as 1/(1+x). This conversion embodies the deep bridge between discrete summation and continuous integration. As a final plausibility check, every term in the sum is positive and bounded above, the partial sums increase steadily toward a finite limit, and the value \pi/4 \approx 0.785 lies plausibly between the magnitudes of the first and last terms for large n, so both the analytic conversion and the numerical intuition agree on the answer.
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- definite integral as limit of sum
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
π/4
A limit of a sum indexed from 1 to n with a 1/n scaling factor is the signature of the Riemann-sum definition, which lets us replace a hard discrete sum by a continuous definite integral. The strategy is to factor out 1/n as the sample spacing and identify k/n as the sample point, so we rewrite \sum_{k=1}^{n}\frac{n}{n^2+k^2} = \sum_{k=1}^{n}\frac{1}{n}\cdot\frac{1}{1 + (k/n)^2}. As n \to \infty, setting x = k/n and dx = 1/n, the sum converges to \int_0^1 \frac{dx}{1 + x^2} = [\arctan x]_0^1 = \arctan 1 - \arctan 0 = \pi/4. Option \pi/2 would arise from integrating over [0, \infty) instead of the unit interval. Option 1 ignores the arctangent structure of the integrand. Option \ln 2 corresponds to a different integrand such as 1/(1+x). This conversion embodies the deep bridge between discrete summation and continuous integration. As a final plausibility check, every term in the sum is positive and bounded above, the partial sums increase steadily toward a finite limit, and the value \pi/4 \approx 0.785 lies plausibly between the magnitudes of the first and last terms for large n, so both the analytic conversion and the numerical intuition agree on the answer.
This hard difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integral as limit of sum. It appeared in the 2025 exam.
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