Infinite Series practice questions — AP Calculus BC

Here $a = 3$ and $r = \tfrac{1}{2}$, so $|r| < 1$.

$$\sum_{n=0}^{\infty} 3\left(\tfrac{1}{2}\right)^{n} = \dfrac{a}{1-r} = \dfrac{3}{1 - \tfrac{1}{2}} = \dfrac{3}{\tfrac{1}{2}} = 6.$$

At $n = 4$ the coefficient is $\dfrac{1}{4!} = \dfrac{1}{24}$.

(Quick check: $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$.)

$$\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\dfrac{|x|}{n+1} = 0 < 1 \quad \text{for every } x.$$

The series converges for all $x \in \mathbb{R}$, so $R = \infty$ (this is the Maclaurin series of $e^{x}$).

For (c): $\lim_{n\to\infty} \dfrac{n}{2n+1} = \dfrac{1}{2} \ne 0$, so the series diverges by the $n$-th term test.

For (a), (b), (d) the term limits are all $0$, so the test is inconclusive (and (d), the harmonic series, actually diverges by other means).

For all $n \ge 1$, $0 < \dfrac{1}{n^{2}+1} \le \dfrac{1}{n^{2}}$.

$\sum \dfrac{1}{n^{2}}$ is a convergent $p$-series ($p = 2 > 1$).

By direct comparison, $\sum \dfrac{1}{n^{2}+1}$ also converges.

$$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{1/(n+1)!}{1/n!} = \dfrac{n!}{(n+1)!} = \dfrac{1}{n+1}.$$

Then $L = \displaystyle\lim_{n\to\infty} \dfrac{1}{n+1} = 0 < 1$, so the series converges absolutely.