We will first use the Ratio Test: lim n→∞ |an+1/an|
lim n→∞ |[(n+1)!(4x-1)n+1]/[n!(4x-1)n]|
Note: (n+1)! can be written as (n+1)n! which will be helpful in cancelling. Also recall we can turn (4x-1)n+1 into (4x-1)n(4x-1)
Employing both of these tactics gives lim n→∞ |[(n+1)n!(4x-1)n(4x-1)]/[n!(4x-1)n]|
We can now cancel n! and (4x-1)n to get lim n→∞ |(n+1)(4x-1)|
Notice our limit deals with n so we can take the stuff with x to make things a little simpler, but do not forget anything we take out needs to have absolute values!
|4x-1| lim n→∞ |(n+1)|
The limit evaluates to infinity and by the Ratio Test criteria for convergence we must set everything <1
Because of this infinity (we can think of it as "dividing both sides by infinity" and 1/∞=0) we get that our radius is 0. If the radius is 0 then we have no interval, just a single point.
If we take the 4x-1 and set it equal to zero we find that x=1/4.
Thus, the series on convergent at this point
Hope this helps! Reading such a complex problem may be difficult to understand, so please feel free to message me and we can set up a quick session to go over any clarifications!
