Does the sequence $\sum_n{n^2\sin (\pi/2^n)}$ converge or diverge? Does the sequence $\sum_n{n^2\sin (\pi/2^n)}$ converge or diverge? I have to use the comparison criteria for limits
 A: For all $x>0$, we have $\sin x < x$. Thus, $\sin\left(\pi/2^n\right)<\pi/2^n$.
Does that give you enough of a start to use the comparison test?
A: Notice that all the terms of the series are positive, therefore it makes sense to try to use the limit comparison test:
$$\lim _{n \to \infty} \frac {n^2 \sin \frac \pi {2^n}} {\frac {n^2} {2^n}} = \lim _{n \to \infty} \frac {\sin \frac \pi {2^n}} {\frac 1 {2^n}} = \lim _{t \to 0} \frac {\sin \pi t} t = \pi \in (0, \infty)$$
which shows that your series has the same behaviour as $\sum \frac {n^2} {2^n}$. For this series (which, too, has positive terms), either the root test, or the ratio test are enough. Using the root test (and the fact that $\sqrt[n] n \to 1$),
$$\lim _{n \to \infty} \sqrt[n] {\frac {n^2} {2^n}} = \lim _{n \to \infty} \frac {\sqrt[n] n^2} 2 = \frac 1 2 < 1$$
which means that the last series is convergent, hence the given series will be convergent too.
A: By the positive  limit comparison test,
$$n^2\sin(\frac{\pi}{2^n})\sim n^2\frac{\pi }{2^n}\;\;\;(n\to+\infty)$$
and by ratio test
$$\lim_{n\to+\infty}\frac{\pi \frac{(n+1)^2}{ 2^{n+1}}}{\pi \frac{n^2}{2^n  }}=\frac{1}{2}<1$$
the series converges.
A: $\lim_{n\to \infty} n^2\sin(\frac{\pi}{2^n})$
$=\lim_{n\to \infty} \{\dfrac{\sin (\frac{\pi}{2^n})}{(\frac{\pi}{2^n})}\}\times (\frac{\pi}{2^n})\times n^2$
$=\lim_{n\to \infty} \pi \dfrac{n^2}{2^n}= 0$
Now compare the series with $\sum \frac{\pi}{2^n}$ which converges and so the given series converges
