Stephanos T. answered • 04/26/20

Experienced Mathematics Tutor With a Bachelors in Mathematics

We will first use the Ratio Test: lim n→∞ |a_{n+1}/a_{n}|

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!