Sum of sines!

Calculus Level 4

lim ⁡ n → ∞ ∑ k = 1 n sin ⁡ ( k n 2 ) = ? \large\lim_{n\to\infty} \sum_{k=1}^n \sin\left( \dfrac {k}{n^2}\right) = \, ?


The answer is 0.5.

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Abdelghani Ssoris by Abdelghani ßoris , 24, Morocco

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2 solutions

We have:

And we can demonstrate that; :

So:

And;

Therfore:

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Otto Bretscher
Jan 9, 2016

The formula for the sum of sines in an arithmetic progression gives

∑ k = 1 n sin ⁡ ( k n 2 ) = sin ⁡ ( n + 1 2 n 2 ) sin ⁡ ( 1 2 n ) sin ⁡ ( 1 2 n 2 ) \sum_{k=1}^{n}\sin\left(\frac{k}{n^2}\right)=\frac{\sin(\frac{n+1}{2n^2})\sin(\frac{1}{2n})}{\sin(\frac{1}{2n^2})}

Multiplying numerator and denominator with n 2 n^2 and using lim ⁡ x → 0 sin ⁡ x x = 1 \lim_{x\to 0}\frac{\sin{x}}{x}=1 , we find a limit of ( 0.5 ) ( 0.5 ) 0.5 = 0.5 \frac{(0.5)(0.5)}{0.5}=\boxed{0.5}

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You can just replace sin ⁡ ( k n 2 ) \sin (\frac{k}{n^2 } ) by k n 2 \frac{k}{n^2} , as we have the fact that lim ⁡ x → 0 sin ⁡ x x = 1 \lim_{x\to 0} \frac{\sin x }{x} =1 .

Hasan Kassim - 5 years, 5 months ago

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That step would require a careful justification. You can't just "replace" sin ⁡ x \sin{x} by x x in any limit you see. My strategy is to find the sum first and then "drop the sines" (now the step is easy to justify).

Otto Bretscher - 5 years, 5 months ago

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