Definite Integral As Limit Of Sum

Question

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.

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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 	o \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.

Definite Integral As Limit Of Sum

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