lim _{n rightarrow infty}left(frac{1}{1^2+n^2 | vedclass

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Question

(B) The given expression can be written as a summation: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r}{r^2+n^2}$ Divide the numerator and d

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given expression can be written as a summation: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r}{r^2+n^2}$ Divide the numerator and denominator by $n^2$: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r/n^2}{(r/n)^2+1} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r/n}{(r/n)^2+1}$ This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $x = r/n$ and $dx = 1/n$: $S = \int_{0}^{1} \frac{x}{x^2+1} dx$ Let $u = x^2+1$,then $du = 2x dx$,so $x dx = du/2$: $S = \frac{1}{2} \int_{1}^{2} \frac{1}{u} du = \frac{1}{2} [\log |u|]_{1}^{2}$ $S = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$

$\lim _{n \rightarrow \infty}\left(\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\ldots+\frac{n}{n^2+n^2}\right)=$

Similar Questions

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1^2}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots \left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=$

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If $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$,then $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ is equal to :

$\lim _{n \rightarrow \infty}\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{n+2 n}{n^2+(2n)^2}\right]=$

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