Determine convergence or divergence for this series ? sin k/(k+1)!

1 Answer
Jul 28, 2018

The series is absolutely convergent.

Explanation:

We know that:

abs (sinx) <=1|sinx|1 for any x in RR

Then:

abs (sink)/((k+1)!) <= 1/((k+1)!)

Now we can prove that the series:

sum_(n=0)^oo 1/((k+1)!)

is convergent using the ratio test, as:

lim_(k->oo) (1/((k+2)!))/(1/((k+1)!)) = lim_(k->oo) ((k+1)!)/((k+2)!) = lim_(k->oo) 1/(k+2) = 0

and then by direct comparison also the series:

sum_(n=0)^oo abs (sink)/((k+1)!)

is convergent,