Limits 3

Calculus Level 4

lim ⁡ n → ∞ 1 n ∑ r = 1 2 n r n 2 + r 2 \large \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{2n} \dfrac{r}{\sqrt{n^{2}+r^{2}}} If the limit above exists, find it to three decimal places.


The answer is 1.236.

Divyansh Tripathi by divyansh tripathi , 21, India

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1 solution

We can express the function as f ( r n ) = r n ( r n ) 2 + 1 \displaystyle f(\frac{r}{n})=\frac{\frac{r}{n}}{\sqrt{(\frac{r}{n})^2+1}} , So using Riemann Sums

S = lim ⁡ n → ∞ ∑ r = 1 2 n f ( r n ) \displaystyle S = \lim_{n\to\infty}\sum_{r=1}^{2n} f(\frac{r}{n}) is equivalent to : S = ∫ 0 2 x 1 + x 2 d x \displaystyle S = \int_{0}^{2}\frac{x}{\sqrt{1+x^2}}dx

Substitute x 2 = t x^2=t & S = [ 1 + t ] 0 4 = 5 − 1 ≈ 1.236 \displaystyle S = [\sqrt{1+t}]_{0}^{4} = \sqrt{5}-1\approx 1.236

Evaluating we get S = 1.236 S=\boxed{1.236}

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