The value of lim _{n rightarrow infty} sum_{r | vedclass

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Question

(C) We express the sum as a definite integral using the limit of a sum formula: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4} = \lim

Vedclass · Accepted Answer

$\frac{1}{4} \log _{e} 2$

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(C) We express the sum as a definite integral using the limit of a sum formula: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n^3(r/n)^3}{n^4((r/n)^4+1)} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(r/n)^3}{(r/n)^4+1}$.Let $x = r/n$, then as $n \rightarrow \infty$, the sum becomes the integral $\int_0^1 \frac{x^3}{x^4+1} dx$.To solve this, let $u = x^4+1$, then $du = 4x^3 dx$, or $x^3 dx = \frac{1}{4} du$.The integral becomes $\frac{1}{4} \int_1^2 \frac{1}{u} du = \frac{1}{4} [\log_e u]_1^2$.$= \frac{1}{4} (\log_e 2 - \log_e 1) = \frac{1}{4} \log_e 2$.

The value of $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4}$ is

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