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

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Apr 13, 2015

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.

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