Question

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

Answer 1

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:

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.