Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$
Options:
Solution:
2422 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}+\frac{3}{2} \ln 2$
We have,
$\begin{gathered}\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots \frac{2}{n}\right] \\ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{1+\frac{3 r}{n}}{1+\left(\frac{r}{n}\right)^2}\right)=\int_0^1\left(\frac{1+3 x}{1+x^2}\right) d x \\ =\int_0^1 1\left(\frac{1}{1+x^2}+\frac{3 x}{1+x^2}\right) d x \\ =\left[\tan ^{-1} x+\frac{3}{2} \log \left(1+x^2\right)\right]_0^1 \\ =\tan ^{-1} 1+\frac{3}{2} \log 2=\frac{\pi}{4}+\frac{3}{2} \log 2\end{gathered}$
$\begin{gathered}\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots \frac{2}{n}\right] \\ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{1+\frac{3 r}{n}}{1+\left(\frac{r}{n}\right)^2}\right)=\int_0^1\left(\frac{1+3 x}{1+x^2}\right) d x \\ =\int_0^1 1\left(\frac{1}{1+x^2}+\frac{3 x}{1+x^2}\right) d x \\ =\left[\tan ^{-1} x+\frac{3}{2} \log \left(1+x^2\right)\right]_0^1 \\ =\tan ^{-1} 1+\frac{3}{2} \log 2=\frac{\pi}{4}+\frac{3}{2} \log 2\end{gathered}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.