Limit problem

Calculus Level 3

If L = lim ⁡ n → ∞ 1 n ∑ j = 1 n cos ⁡ ( j π 2 n ) L=\displaystyle\lim \limits_{n\to \infty }\frac{1}{n} \displaystyle\sum \limits^{n}_{j=1}\cos \left( \frac{j\pi }{2n} \right) find ⌊ 1 L ⌋ \left\lfloor \frac{1}{L}\right\rfloor .


The answer is 1.

Refaat M. Sayed by Refaat M. Sayed , 24, Egypt

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2 solutions

Rishabh Jain
Jan 27, 2016

Shouldn't the answer be 0 since ⌊ 2 π ⌋ = 0 \lfloor \dfrac{2}{\pi} \rfloor=0

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I agree too!

Rudraksh Shukla - 5 years, 4 months ago

No.. 1 is the correct answer.
L = 2 π L = \dfrac{2}{\pi}

Akhil Bansal - 5 years, 4 months ago

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The question has been rectified ..Previously question was about finding ⌊ L ⌋ \left\lfloor L \right\rfloor which is obviously 0.

Rishabh Jain - 5 years, 4 months ago
Amodh Makhija
Jan 27, 2016

Yes, the answer should have been [ 2/π ]= 0 \boxed{0}

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