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UPSC 2018 Maths Optional Paper 1 Q1d — Step-by-Step Solution 10 marks · Section A
Riemann Definite Integral; Integrability · Calculus · Read the full method →
Question
Find the limit lim n → ∞ 1 n 2 ∑ r = 0 n − 1 n 2 − r 2 \displaystyle\lim_{n\to\infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\sqrt{n^2-r^2} n → ∞ lim n 2 1 r = 0 ∑ n − 1 n 2 − r 2
Technique
Limit of a sum → \to → ∫ 0 1 1 − x 2 d x = π / 4 \int_0^1\sqrt{1-x^2}\,dx=\pi/4 ∫ 0 1 1 − x 2 d x = π /4
Solution
Step 1 — Recognise a Riemann sum
Factor n n n
1 n 2 ∑ r = 0 n − 1 n 2 − r 2 = 1 n 2 ∑ r = 0 n − 1 n 1 − ( r n ) 2 = 1 n ∑ r = 0 n − 1 1 − ( r n ) 2 . \frac1{n^2}\sum_{r=0}^{n-1}\sqrt{n^2-r^2}=\frac1{n^2}\sum_{r=0}^{n-1}n\sqrt{1-\Big(\tfrac rn\Big)^2}=\frac1n\sum_{r=0}^{n-1}\sqrt{1-\Big(\tfrac rn\Big)^2}. n 2 1 r = 0 ∑ n − 1 n 2 − r 2 = n 2 1 r = 0 ∑ n − 1 n 1 − ( n r ) 2 = n 1 r = 0 ∑ n − 1 1 − ( n r ) 2 . This is a Riemann sum for f ( x ) = 1 − x 2 f(x)=\sqrt{1-x^2} f ( x ) = 1 − x 2 [ 0 , 1 ] [0,1] [ 0 , 1 ] x r = r / n x_r=r/n x r = r / n Δ x = 1 / n \Delta x=1/n Δ x = 1/ n
Step 2 — Pass to the integral
Since f ( x ) = 1 − x 2 f(x)=\sqrt{1-x^2} f ( x ) = 1 − x 2 [ 0 , 1 ] [0,1] [ 0 , 1 ]
lim n → ∞ 1 n ∑ r = 0 n − 1 1 − ( r n ) 2 = ∫ 0 1 1 − x 2 d x . \lim_{n\to\infty}\frac1n\sum_{r=0}^{n-1}\sqrt{1-\Big(\tfrac rn\Big)^2}=\int_0^1\sqrt{1-x^2}\,dx. n → ∞ lim n 1 r = 0 ∑ n − 1 1 − ( n r ) 2 = ∫ 0 1 1 − x 2 d x . Step 3 — Evaluate the integral
∫ 0 1 1 − x 2 d x \int_0^1\sqrt{1-x^2}\,dx ∫ 0 1 1 − x 2 d x π 4 \dfrac\pi4 4 π ∫ 1 − x 2 d x = 1 2 [ x 1 − x 2 + arcsin x ] \int\sqrt{1-x^2}\,dx=\tfrac12\!\left[x\sqrt{1-x^2}+\arcsin x\right] ∫ 1 − x 2 d x = 2 1 [ x 1 − x 2 + arcsin x ] 0 0 0 1 1 1 1 2 arcsin 1 = π 4 \tfrac12\arcsin1=\tfrac\pi4 2 1 arcsin 1 = 4 π
Answer
lim n → ∞ 1 n 2 ∑ r = 0 n − 1 n 2 − r 2 = π 4 . \boxed{\;\lim_{n\to\infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\sqrt{n^2-r^2}=\frac{\pi}{4}.\;} n → ∞ lim n 2 1 r = 0 ∑ n − 1 n 2 − r 2 = 4 π .