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Lim n→∞Σr=1^n(r^2/(n^3+r^3 equals Single correct option
lim n → ∞ ∑ r = 1 n r 2 n 3 + r 3 \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\dfrac{r^2}{n^3+r^3} n → ∞ lim r = 1 ∑ n n 3 + r 3 r 2 equals
A. 1 3 ln 2 \dfrac13\ln 2 3 1 ln 2 B. ln 2 \ln 2 ln 2 C. 1 3 \dfrac13 3 1 D. 2 3 ln 2 \dfrac23\ln 2 3 2 ln 2 Show answer and solution Answer: (A) 1 3 ln 2 \dfrac13\ln 2 3 1 ln 2
Divide numerator and denominator by
n 3 n^3 n 3 :
r 2 n 3 + r 3 = 1 n ⋅ ( r / n ) 2 1 + ( r / n ) 3 \frac{r^2}{n^3+r^3}=\frac1n\cdot\frac{(r/n)^2}{1+(r/n)^3} n 3 + r 3 r 2 = n 1 ⋅ 1 + ( r / n ) 3 ( r / n ) 2 . So the sum is a Riemann sum of
f ( x ) = x 2 1 + x 3 f(x)=\frac{x^2}{1+x^3} f ( x ) = 1 + x 3 x 2 over
[ 0 , 1 ] [0,1] [ 0 , 1 ] with
x = r n x=\frac rn x = n r . The limit equals
∫ 0 1 x 2 1 + x 3 d x \int_0^1\frac{x^2}{1+x^3}\,dx ∫ 0 1 1 + x 3 x 2 d x . Put
u = 1 + x 3 u=1+x^3 u = 1 + x 3 ,
d u = 3 x 2 d x du=3x^2dx d u = 3 x 2 d x : the integral is
1 3 ∫ 1 2 d u u = 1 3 ln 2 \frac13\int_1^2\frac{du}{u}=\frac13\ln2 3 1 ∫ 1 2 u d u = 3 1 ln 2 . A numerical sum with
n = 2 × 10 5 n=2\times10^5 n = 2 × 1 0 5 gives
0.23105 0.23105 0.23105 , equal to
ln 2 3 \frac{\ln2}{3} 3 l n 2 .
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