How do you test for convergence for #(sin 4n)/(4^n)# for n=1 to infinity?

Question

Elijah Abram · Accepted Answer

Since the function sinus is with range #[-1,1]#, than: #sin4n<=1# and so: #sin(4n)/4^n<=1/4^n<=1/n^2# (for #n# big enough) that is a convergent series.  So our series is convergent for the principle of comparision.

Christopher Allers · Answer

undefined To test for convergence of the series $ \frac{\sin(4n)}{4^n} $ from $ n = 1 $ to $ \infty $, we can use the limit comparison test. Let's consider a series $ \sum_{n=1}^{\infty} a_n $ and another series $ \sum_{n=1}^{\infty} b_n $. If $ \lim_{n \to \infty} \frac{a_n}{b_n} $ is a finite positive number, then either both series converge or both series diverge.

In this case, we'll compare $ \frac{\sin(4n)}{4^n} $ to $ \frac{1}{4^n} $ because $ |\sin(4n)| \leq 1 $.

So, we'll consider the series $ \sum_{n=1}^{\infty} \frac{1}{4^n} $. This is a geometric series with common ratio $ r = \frac{1}{4} $. It converges since $ |r| < 1 $.

Now, we'll compute the limit:

$$ \lim_{n \to \infty} \frac{\frac{\sin(4n)}{4^n}}{\frac{1}{4^n}} = \lim_{n \to \infty} \sin(4n) $$

Since $ |\sin(4n)| \leq 1 $ for all $ n $, and $ \sin(4n) $ oscillates between $ -1 $ and $ 1 $, the limit is bounded between $ -1 $ and $ 1 $, which means it's finite.

Therefore, by the limit comparison test, since $ \sum_{n=1}^{\infty} \frac{1}{4^n} $ converges, $ \sum_{n=1}^{\infty} \frac{\sin(4n)}{4^n} $ also converges.