Question

$\large \displaystyle \sf \sum \limits_{k = 2}^{ \infty} \frac{1}{ {2}^{k - 2} } \tan \bigg( \frac{\pi}{ {2}^{k} } \bigg )$​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

We can use the formula for the sum of an infinite geometric series to simplify the given sum:

$\begin{align}\sum_{k=2}^\infty \frac{1}{2^{k-2}} \tan \left(\frac{\pi}{2^k}\right) &= \sum_{k=2}^\infty \frac{1}{2^{k-2}} \cdot \frac{\sin \left(\frac{\pi}{2^k}\right)}{\cos \left(\frac{\pi}{2^k}\right)} \\&= \sum_{k=2}^\infty \frac{\sin\left(\frac{\pi}{2^k}\right)}{2^{k-2} \cos \left(\frac{\pi}{2^k}\right)} \\&= \sum_{k=2}^\infty \frac{\sin \left(\frac{\pi}{2^k}\right)}{2^{k-1} \sin \left(\frac{\pi}{2^k}\right) \cos \left(\frac{\pi}{2^k}\right)} \\&= \sum_{k=2}^\infty \frac{1}{2^{k-1} \cos \left(\frac{\pi}{2^k}\right)^2} \\\end{align}$

$\rm Now, we \: can \: use \: the identity \\ \cos(2\theta) = 2\cos^2 \theta - 1 \\ \rm to \: simplify \: the \: denominator:$

$\begin{align}\cos \left(\frac{\pi}{2^k}\right)^2 &= \frac{1 + \cos \left(\frac{\pi}{2^{k-1}}\right)}{2} \\&= \frac{1 + \cos \left(\frac{\pi}{2^{k-2}}\right)}{2^2} \\&= \frac{1 + \cos \left(\frac{\pi}{2^{k-3}}\right)}{2^3} \\& \vdots \\&= \frac{1 + \cos \left(\pi\right)}{2^k} \\&= \frac{2}{2^k} \\&= \frac{1}{2^{k-1}} \\\end{align}$

Substituting this back into the sum, we get:

$\begin{align}\sum_{k=2}^\infty \frac{1}{2^{k-1} \cos \left(\frac{\pi}{2^k}\right)^2} &= \sum_{k=2}^\infty \frac{1}{2^{k-1} \cdot \frac{1}{2^{k-1}}} \\&= \sum_{k=2}^\infty 1 \\&= \infty\end{align}$

Therefore, the given sum diverges to infinity.