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

1 Answer
Jul 2, 2015

By comparing it with #1/n#

Explanation:

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:
enter image source here

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.