Find the limit

Calculus Level 4

lim n k = 1 n n k 2 + n 2 \large\lim_{n \to \infty}\sum_{k=1}^{n}\dfrac{n}{k^2+n^2}

The limit above has a closed form. Find this closed form.

Give your answer to 3 decimal places.


The answer is 0.785398.

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1 solution

Rishabh Jain
Apr 30, 2016

Let T \mathfrak T denote the given limit. T = lim n k = 1 n n k 2 + n 2 = lim n 1 n k = 1 n 1 ( k n ) 2 + 1 \Large \mathfrak {T}=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{n}{k^2+n^2}\\\Large=\lim_{n \to \infty}\dfrac 1n\sum_{k=1}^{n}\dfrac{1}{\left(\color{#D61F06}{\dfrac{k}{n}}\right)^2+1}

By Reimann Sums ,

T = 0 1 d x x 2 + 1 \Large\mathfrak T=\displaystyle\int_0^1\dfrac{\mathrm{d}x}{\color{#D61F06}{x^2}+1} = tan 1 x 0 1 = tan 1 ( 1 ) \Large=\left|\tan^{-1}x\right|_0^{1}=\tan^{-1}(1) = π / 4 0.785398 \Large=\pi/4\approx \boxed{0.785398}

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