The value of limn-6n-2-6n-22-6n- -2n-1-6n - 43- is

Lim_n→∞ nbsp;π6n[^2(π6n)+^2(2·π6n)+⋯ +^2((n 1)π6n) + 43] = π6lim_n→∞ nbsp;1n∑_r=1^n^2(r·π6n) =π6∫_0^1^2(π x6) dx =[tanπ x6]_0^1 =1√(3)=√(3)3 ...

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Definite Integral as Limit of Sum

The value of ...

Question

The value of $lim n \to \infty \frac{π}{6 n} [{sec}^{2} (\frac{π}{6 n}) + {sec}^{2} (2 \cdot \frac{π}{6 n}) + \dots + {sec}^{2} ((n - 1) \frac{π}{6 n}) + \frac{4}{3}]$ is

\frac{\sqrt{3}}{3}

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\sqrt{3}

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2

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\frac{2}{\sqrt{3}}

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Similar questions

lim n \to \infty {(tan (\frac{π}{2 n}) \cdot tan (\frac{2 π}{2 n}) \dots tan (\frac{(n - 1) π}{2 n}))}^{1 / n}

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A_{r} = cos \frac{π}{3^{r}} + i sin \frac{π}{3^{r}},

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sin \frac{π}{14} \cdot sin \frac{3 π}{14} \cdot sin \frac{5 π}{14} \cdot sin \frac{7 π}{14} \cdot sin \frac{9 π}{14} \cdot sin \frac{11 π}{14} \cdot sin \frac{13 π}{14}

c o s \frac{π}{33} \cdot c o s \frac{2 π}{65} \cdot \frac{4 π}{65} \cdot c o s \frac{8 π}{65} \cdot c o s \frac{16 π}{65} \cdot c o s \frac{32 π}{65}

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