Does This Mathematical Series Converge?

[twoflower]

Hi all,

I have this series:

<br /> \sum_{n = 1}^{+\infty} \tan \left( \frac{\pi}{4^{n}} \right) . \sin 2^{n}<br />

I have to find out whether it converges or not, but I don't know how should I start. The only idea coming to my mind is to use Abel-Dirichlet's rule for convergence, but I don't know how to prove that sin has limited partial sums. Then I could use the rule I hope.

Or is there any other and more clever way how to prove the convergence?

Thank you.

 

[vincentchan]

Hints
<br /> \sum_{n = 1}^{+\infty} \tan \left( \frac{\pi}{4^{n}} \right) \cdot \sin 2^{n} &lt; \sum_{n = 1}^{+\infty} \tan \left( \frac{\pi}{4^{n}} \right)<br />

 

[wais]

Hello!

To determine the convergence of this series, you can use the comparison test. First, note that

\tan \left( \frac{\pi}{4^{n}} \right) = \frac{\sin \left( \frac{\pi}{4^{n}} \right)}{\cos \left( \frac{\pi}{4^{n}} \right)},

and since \sin \left( \frac{\pi}{4^{n}} \right) is always between -1 and 1, we can rewrite the series as

\sum_{n = 1}^{+\infty} \frac{\sin \left( \frac{\pi}{4^{n}} \right)}{\cos \left( \frac{\pi}{4^{n}} \right)} \cdot \sin 2^{n}.

Now, since \cos \left( \frac{\pi}{4^{n}} \right) \to 1 as n \to \infty, we can use the comparison test with the series \sum_{n = 1}^{+\infty} \sin 2^{n}. This series is known to be convergent, as it is a geometric series with a common ratio of 2. Therefore, by the comparison test, the original series also converges.

Hope this helps!