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

How do you solve the series #sin (1/n)# using comparison test?

Answer 1

By comparing it with #1/n#

The sine function has this weird property that for very small values of #x#: #sin(x) = x#
You can see this easily by plotting the graph for #y = sin(x)# and the graph for #y=x# over each other:

You can see that when #x->0#, #sinx=x#

So this also means that for very small values of #1/n#, #sin(1/n)=1/n#
When does #1/n# become very small? When #n# is very big, like infinity. So, at infinity we can compare #sin(1/n)# with #1/n#.
We also know that #1/n# diverges at infinity, so #sin(1/n)# must also diverge at infinity.

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Answer 2

Charles Arocha

To solve the series ( \sin\left(\frac{1}{n}\right) ) using the comparison test:

Choose a series ( b_n ) that is easier to evaluate and for which it is known whether it converges or diverges.
Establish a comparison between the given series ( \sin\left(\frac{1}{n}\right) ) and the chosen series ( b_n ).
Show that ( | \sin\left(\frac{1}{n}\right) | \leq | b_n | ) for all ( n ) beyond some index ( N ).
Use the convergence or divergence of the series ( b_n ) to determine the convergence or divergence of ( \sin\left(\frac{1}{n}\right) ).

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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.