Mathematics · 2019
JEE Main · 12 January 2019, Shift 2 · Q76
lim_n → ∞(n/(n^2+1^2)+n/(n^2+2^2)+n/(n^2+3^2)+…+1/(5 n)) is equal to:
$$\lim _{n \rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{1}{5 n}\right)
$$
is equal to:
Official answer
(1)
$\displaystyle \tan ^{-1}(2)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.