lim _{n rightarrow infty} sum_{k=1}^n frac{k} | vedclass

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(A) We have,$I = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k/n}{1+(k/n

Vedclass · Accepted Answer

$\frac{1}{2} \log 2$

Vedclass · Answer

(A) We have,$I = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k/n}{1+(k/n)^2}$ By the definition of definite integral as a limit of a sum, we replace $\frac{k}{n}$ with $x$ and $\frac{1}{n}$ with $dx$, where the limits of integration are from $0$ to $1$. $I = \int_0^1 \frac{x}{1+x^2} dx$ Let $u = 1+x^2$, then $du = 2x dx$, which implies $x dx = \frac{1}{2} du$. When $x=0, u=1$ and when $x=1, u=2$. $I = \frac{1}{2} \int_1^2 \frac{1}{u} du = \frac{1}{2} [\log |u|]_1^2 = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$.

$\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = $

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