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Compute $\lim\limits_{n\to \infty} n \sqrt{2^{n-1}} \cos^{n-1} \alpha$, where $\alpha \in \left(\frac{\pi} {4}, \frac{\pi} {2} \right) $.
I think that we may rewrite it as an $\frac{\infty} {\infty} $ indeterminate form and use L'Hopital's rule, but this looks messy and I think there should be some better way to do it.

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The answer is 0.

  1. For $\alpha \in (\frac{\pi}{4}, \frac{\pi}{2})$, the value of $\cos \alpha$ is a fixed nonegative number $a$ strictly less than $\frac{1}{\sqrt{2}}$. [Indeed $\cos \frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$ and $\cos(\cdot)$ is montonically decreasing on the interval $[\frac{\pi}{4}, \frac{\pi}{2})$. ]

  2. So for each positive integer $n$ and $a = \cos \alpha$; $\alpha \in (\frac{\pi}{4}, \frac{\pi}{2})$, the value of $\sqrt{2^{n-1}}a^{n-1} = b^{n-1}$ for some fixed nonegative $b$ less than 1 [make sure you see why].

  3. $\lim_{n \rightarrow \infty} nb^{n-1}$ is 0 for any fixed nonegative $b < 1$.

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Hint: This expression is of the form $$\sum_{n=1}^\infty nx^{n-1}$$ This looks like the derivative of something! Do you see it?

EDIT: This expression is not quite in that form, but you may be able to say that since we know this converges for $x\in (-1,1)$ that the limit of the summand is $0$.

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    $\begingroup$ Isn't the summation index $n $? Also why did you take summation? $\endgroup$ – Thomas Shelby Mar 11 at 18:19
  • $\begingroup$ That is the derivative of $x^n$, but how do we use this and why is it a series? $\endgroup$ – MathEnthusiast Mar 11 at 18:21
  • $\begingroup$ I'm not seeing this either. $\endgroup$ – Mike Mar 11 at 18:22
  • $\begingroup$ Sorry, I thought you were taking the limit of the sum. Ah well. Technically this could be salvaged by saying that this is part of the tail of the series, whose limit must be 0. $\endgroup$ – Isaac Browne Mar 11 at 18:22

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