Infinite Series practice questions — AP Calculus BC

Here $a=3$ and $r=\frac{1}{2}$, so $\mathrm{\mid }r\mathrm{\mid }<1$.

$$\sum _{n=0}^{\mathrm{\infty }}3{\left(\frac{1}{2}\right)}^n={\displaystyle \frac{a}{1-r}}={\displaystyle \frac{3}{1-\frac{1}{2}}}={\displaystyle \frac{3}{\frac{1}{2}}}=6.$$

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

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

$$\underset{n\to \mathrm{\infty }}{\lim }\mid {\displaystyle \frac{a_{n+1}}{a_n}}\mid =\underset{n\to \mathrm{\infty }}{\lim }{\displaystyle \frac{\mathrm{\mid }x\mathrm{\mid }}{n+1}}=0<1\text{for every }x\mathrm{.}$$

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

For (c): ${\lim }_{n\to \mathrm{\infty }}{\displaystyle \frac{n}{2n+1}}={\displaystyle \frac{1}{2}}\mathrm{\ne }0$, 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 $n\ge 1$, $0<{\displaystyle \frac{1}{n^2+1}}\le {\displaystyle \frac{1}{n^2}}$.

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

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

$$\mid {\displaystyle \frac{a_{n+1}}{a_n}}\mid ={\displaystyle \frac{1\mathrm{/}(n+1)!}{1\mathrm{/}n!}}={\displaystyle \frac{n!}{(n+1)!}}={\displaystyle \frac{1}{n+1}}\mathrm{.}$$

Then $L={\displaystyle \underset{n\to \mathrm{\infty }}{\lim }{\displaystyle \frac{1}{n+1}}=0<1}$, so the series converges absolutely.