Question

How do you test for convergence of `Sigma (3n-7)/(10n+9)` from `n=[0,oo)`?

Answer 1

By putting an limit in front.

Explanation:

`lim_(n=>oo)((3n-7)/(10n+9))`
In this case, if the result is `!=0` there's divergence.
But the result is something that we cannot solve.
`=>[oo/oo]`
So we are going to isolate n in the equation:
`=>lim_(n=>oo)((n(3-7/n))/(n(10+9/n)))`
`=lim_(n=>oo)((3-7/n)/(10+9/n))`
And we just have to solve it!
`=(3-7/oo)/(10+9/oo)`
`=(3-0)/(10+0)`
`=3/10`
The result is `!=0` so there's divergence!
(I'not an english speaker so I don't know the name of this method. Feel free to add it.)