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Test whether the series $\sum_{n=1}^\infty\frac{1+\sin(n)}{2^n}$ is converges or diverges. I used the limit comparison method in order to test its convergence. I chose $a_n = \frac{1+\sin(n)}{2^n}$ and $b_n = \frac{1}{2^n}$.

$$ \lim_{n \rightarrow \infty}\frac{\frac{1+\sin(n)}{2^n}}{\frac{1}{2^n}} = \lim_{n \rightarrow \infty}({1+\sin(n)}) = \infty $$

However, WolframAlpha suggests that the series is in fact convergent. If this is the case, then is the way I set up the limit comparison test incorrect or is there another test that proves this series is convergent? Any help would be appreciated!

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  • $\begingroup$ $$\lim_{n \to \infty} (1+\sin{n}) \neq \infty$$ $\endgroup$ – projectilemotion Mar 25 '17 at 14:06
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    $\begingroup$ simply use $|1+sin(n)|\le2$ $\endgroup$ – zwim Mar 25 '17 at 14:07
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First of all, $\lim_{n \to \infty} 1 + \sin(n)$ is not equal to $\infty$. It doesn't exist, since $\sin(n)$ oscillates between $-1$ and $1$ as $n$ tends to infinity.

Try instead to compare $a_n =\sum_{i=1}^n \frac{1+\sin(n)}{2^n}$ with $b_n = \sum_{i=1}^n\frac{1+1}{2^n} = \sum_{i=1}^n\frac{2}{2^n}$. Note that since $\sin(n) \leq 1$ for all $n$, we have $a_n \leq b_n$ for all $n$. Thus,

$$\lim_{n \to \infty} \sum_{i=1}^n a_n \leq \lim_{n \to \infty} \sum_{i=1}^nb_n$$

Since $\lim_{n\to \infty} \sum_{i=1}^n b_n = \lim_{n \to \infty} \sum_{i=1}^n\frac{2}{2^n} = 2$, $\lim_{n \to \infty} \sum_{i=1}^n a_n \leq 2$ must converge.

Edit: As pointed out by zwim, you also have to make sure that $\sum a_n$ doesn't tend to $-\infty$. To do this, note that since $-1 \leq \sin(n)$ for all $n$, we have

$$ 0 = \sum_{i=1}^n \frac{1-1}{2} \leq \sum_{i=1}^n \frac{1+\sin(n)}{2} = \sum_{i=1}^n a_n .$$

Since $\sum_{i=1}^n a_n$ is bounded from below and from above, it must converge.

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    $\begingroup$ You have to bound it inferiorly by 0 also. $\endgroup$ – zwim Mar 25 '17 at 14:10
  • $\begingroup$ @zwim Forgot about that. Fixed. $\endgroup$ – aras Mar 25 '17 at 14:14
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HINT:

$$-1 \leq \sin(n) \leq 1$$

so that $0 \leq (1+ \sin (n)) \leq 2$ and hence $1 +\sin(n)$ is non-negative for all $n \in \mathbb n$.

At any rate, since $0\leq (1 + \sin n) \leq 2,$ we can use $b_n = \frac{2}{2^n}$, and show that $0\leq a_n \leq b_n$, etc, to prove the series converges by using the comparison test, which you seem very capable of using.

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    $\begingroup$ Technically, $-1 \leq \sin(n) \leq 1$. $\endgroup$ – aras Mar 25 '17 at 14:08
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    $\begingroup$ @aras well $sin(x)=\pm 1\iff x=\pm\frac{\pi}{2}+2k\pi \notin\mathbb N$ so the strict inequalities were ok. But anyway, since we have to loose them for the limit process, it does not matter. $\endgroup$ – zwim Mar 25 '17 at 14:15

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