How do you test for convergence #(sin(2n))/(1+(2^n))# from n=1 to infinity?

1 Answer
VNVDVI
Apr 25, 2018

Converges by the Direct Comparison Test.

Explanation:

We can use the Direct Comparison Test for this.

On the interval #[1, oo), -1<=sin(2n)<=1#.

So, for our comparison sequence #b_n,# if we remove #sin(2n)# from the denominator, we get a larger numerator and therefore a larger sequence:

#b_n=1/(1+2^n)#

We can also drop the constant #1# from the denominator. This will not drastically change the behavior of #b_n# in terms of it being larger than #a_n#:

#b_n=1/2^n=(1/2)^n#

Now, we know the series

#sum_(n=1)^oo(1/2)^n# converges as it is a geometric series with the common ratio #|r|=1/2<1#.

Then, since the larger series converges, so must the smaller series.

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