Home > Calculus > Limit Comparison Test for Convergence of an Infinite Series

How do you determine whether the series is convergent or divergent given #sum (sin^2(n))/(n*sqrt(n))# for n=1 to #n=oo#?

Answer 1

Converges by the Comparison Test.

On the interval #[1, oo], sin^2(n)<1# (not #<=1, n ne pi/2+2npi# as we're dealing only with integers).

So, knowing that #sin^2(n)# is always less than one, we can remove it from the sequence #a_n=sin^2n/(nsqrtn)# to create a new sequence

#b_n=1/(nsqrtn)=1/(n(n^(1/2)))=1/n^(3/2)>=a_n# for all #n.# Removing a quantity that is always less than one from the numerator creates a new sequence that is always larger than the previous one.

So, now, we know #sum_(n=1)^oo1/n^(3/2)# is convergent. It is a #p-#series in the form #sum1/n^p# where #p=3/2>1#, so it must converge. We could prove this property by the Integral Test.

Then, since the larger series #sum_(n=1)^oo1/n^(3/2)# converges, so must the smaller series #sum_(n=1)^oo(sin^2n)/(nsqrtn)#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.