$\infty \sum n = 1 \frac{sin (n)}{n!}$ $sum_(n=1)^oo sin(n)/(n!)$ How would i find if it converges or diverges?

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$\infty \sum n = 1 \frac{sin (n)}{n!}$ $sum_(n=1)^oo sin(n)/(n!)$ How would i find if it converges or diverges?

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Answer 1 · 2017-04-13T03:40:10

Note that the function $sin (n)$ $sin(n)$ oscillates in between $- 1 \leq sin (n) \leq 1$ $-1lt=sin(n)<=1$ . This means that when $sin (n)$ $sin(n)$ is multiplied by a function, its absolute value will be less than or equal to the original function.

In this situation, we see that $∣ ∣ ∣ \frac{sin (n)}{n!} ∣ ∣ ∣ \leq \frac{1}{n!}$ $abs(sin(n)/(n!))<=1/(n!)$ .

By the direct comparison test, if we can show that $\frac{1}{n!}$ $1/(n!)$ is a convergent series, then $\frac{sin (n)}{n!}$ $sin(n)/(n!)$ will be a convergent series as well since it's less than (or equal to) $\frac{1}{n!}$ $1/(n!)$ .

We can test the behavior of $\infty \sum n = 1 \frac{1}{n!}$ $sum_(n=1)^oo1/(n!)$ using the ratio test.

Applying the ratio test gives:

$lim_(nrarroo)abs((1/((n+1)!))/(1/(n!)))=lim_(nrarroo)abs((n!)/((n+1)!))=lim_(nrarroo)abs(1/(n+1))=0$

Since $lim_(nrarroo)abs(a_(n+1)/(a_n))=0<1$ , we know that $sum_(n=1)^oo1/(n!)$ is convergent.

Then, as shown before, $sum_(n=1)^oosin(n)/(n!)$ is also convergent through the direct comparison test.

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$\infty \sum n = 1 \frac{sin (n)}{n!}$ $sum_(n=1)^oo sin(n)/(n!)$ How would i find if it converges or diverges?

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$\infty \sum n = 1 \frac{sin (n)}{n!}$ $sum_(n=1)^oo sin(n)/(n!)$ How would i find if it converges or diverges?