Calculus – Conditions for Power Series Convergence at x

Question

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:

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

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

---

From the answer key, the answer is d.

Answer 1

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:

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

2. This is a mistake, related to the typo. Given the convergence at $5$, we can say that $\lim\left|\dfrac{a_{n+1}}{a_n}\right|\le\dfrac12$.