If the Power Series converges at x, which must be true? I'm currently reviewing for tomorrow's Calculus BC exam but I got stuck on this one problem.

$\sum_{n=0}^{\infty} a_n (x-3)^n$ converges at $x = 5$. Which of the following must be true?
a. the series diverges at $x = 0$
b. the series diverges at $x = 1$
c. the series converges at $x = 1$
d. the series converges at $x = 2$
e. the series converges at $x = 6$

My intuition was to plug in $x = 5$ into the power series, resulting in $\sum_{n=0}^{\infty} a_n (2)^n$. From here, I deduced that $a_n < (\frac{1}{2})^n$ as any $\sum_{n=0}^{\infty} x^n$ where $|x| < 1$  converges. However, from here I became lost and I searched for an online solution, leading me to find:

While I understand the use of the ratio test:

*

*Why is the term $a_n$ used on top and of the bottom of the fraction?


*Why is $x = 5$ convergent even though it is not within the $2 < x < 4$ boundary?

From the answer key, the answer is d.
 A: With the given info, the radius of convergence is at least $5-3$ and the series is certainly convergent in $(1,5]$. For other values, convergence is unsure. Hence


*

*a: unsure,

*b: unsure,

*c: unsure,

*d: true,

*e: unsure.
Regarding your specific questions:


*

*This is a typo, the ratio must be $\dfrac{a_{n+1}}{a_n}(x-3)$.

*This is a mistake, related to the typo. Given the convergence at $5$, we can say that $\lim\left|\dfrac{a_{n+1}}{a_n}\right|\le\dfrac12$.
A: This is a power series with radius of convergence greater and equal to 2 (2=5-3).
For A, B, C, D, and E, only D falls in convergence radius (2-3=-1, abs (-1) < 2). It is for sure convergent. For B and C, it is on the "radius". It is not sure. For A and E, it is beyond the "radius". They are even more not sure.  
A: Everything in Yves Daoust's answer above is exactly correct.
I would just add that the handwritten stuff in red in the solution you found online is not so helpful (even after the typo/mistakes are corrected).  However, the handwritten stuff in black is very helpful.  (Although, part III has an important typo ... both 1's should be r's=radius of convergence.) With this correction, the fact that the series converges at x=5 means I is not the case, so either II or III must be the case (as I, II, or III are the only options for power series).  If III is the case, then the smallest the interval of convergence could be is (1,5].
A: Can't we use absolute convergence to also say it must converge at x=1 as well:

ACTUALLY, I just realized my mistake is that I forgot that a_n could be nonpositive.
