I can't seem to find a way to prove the limit of the following. (Note: For clarity this problem is written entirely in degrees) $$ \lim_{n\to\infty}\ \sum_{i = 1}^{n}\arccos\left(\cos^{2}\left(30\right)\cos\left(90 \over n\right) + \sin^{2}\left(30\right)\right) $$ The limit should be about $\color{#555}{\texttt{77.94228634}}$. How would I go about finding the exact limit ?. ( The limit will give the angle displaced by a vector ( b ) fixed at $30$ deg. from another vector ( a ), when ( a ) rotates $90$ deg. And is equal to $\color{#777}{90\cos\left(30\right)}$.
Limit of infinite sum Problem
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1The index is $i$ or $n$? – 2017-01-17
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0@Masacroso I find the limit works out as expected with index $i$. $i$ doesn't show up so the sum has a clear first step simplification. – 2017-01-17
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0Note that each summand is independent of $i$, so in fact we are trying to calculate $\lim_{n\to\infty}n\cdot\arccos\left(\cos^2(30)\cos\left(\frac{90}n\right)+\sin^2(30)\right)$. This can be evaluated using a Taylor expansion. – 2017-01-17
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0@Jason or just L'Hôpital? – 2017-01-17
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0@Masacroso Eh, I guess. The limit still holds with a continuous rather than discrete variable so L'Hopital will work but I am not a big fan of using it where other methods suffice and provide better understanding. – 2017-01-17
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0I still don't see where $i$ is used in the sum. – 2017-01-17
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0I deleted my answer because it dont seem that we can solve this using L'Hôpital. Anyway you cant get $0$ using L'Hôpital, just $0/0$ or $\infty/\infty$ again. – 2017-01-19