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Section 9.1 #29: Determine the convergence or divergence of the sequence with the given $n$th term. If it converges, find the limit:
$$a_n = \dfrac{5}{n+2}.$$
Solution: The sequence converges. Compute
$$\displaystyle\lim_{n \rightarrow \infty} a_n = \displaystyle\lim_{n \rightarrow \infty} \dfrac{5}{n+2} = 0.$$
Section 9.1 #36: Determine the convergence or divergence of the sequence with the given $n$th term. If it converges, find the limit:
$$a_n = \dfrac{5^n}{3^n}.$$
Solution: The sequence diverges. This is because $\dfrac{5}{3} > 1$ and so taking powers of $\dfrac{5}{3}$ makes bigger and bigger numbers exponentially.
Section 9.2 #8: Verify that the series diverges:
$$\displaystyle\sum_{n=0}^{\infty} 4 (-1.05)^n.$$
Solution: The series diverges because it is a geometric series with $r=-1.05$ -- recall that a geometric series converges only when $-1 < r <1$.
Section 9.2 #15: Verify that the series converges:
$$\displaystyle\sum_{n=0}^{\infty} \left( \dfrac{5}{6} \right)^n.$$
Solution: Since $-1 \lt r=\dfrac{5}{6} \lt 1$ we observe that this is a geometric series which is convergent.
Section 9.2 #26: Find the sum of
$$\displaystyle\sum_{n=0}^{\infty} \left( - \dfrac{1}{5} \right)^n.$$
Solution: Since $-1 < r=-\dfrac{1}{5} < 1$, this is a convergent geometric series and its sum is given by the general formula $\displaystyle\sum_{k=a}^{\infty} \dfrac{r^a}{1-r}$ with $a=0$: compute
$$\displaystyle\sum_{n=0}^{\infty} \left( -\dfrac{1}{5} \right)^n = \dfrac{1}{1-(-\frac{1}{5})}= \dfrac{1}{\frac{6}{5}} = \dfrac{5}{6}.$$
Section 9.2 #28: Find the sum of
$$\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)(2n+3)}.$$
Solution: First we apply partial fractions to the summand:
$$\dfrac{1}{(2n+1)(2n+3)} = \dfrac{A}{2n+1} + \dfrac{B}{2n+3}.$$
Multiply by the common denominator to get
$$1 = A(2n+3) + B(2n+1) = (2A+2B)n+(3A+B).$$
Equating coefficients leads us to the following system of two linear equations in two variables:
$$\left\{ \begin{array}{rrll}
2A &+2B &= 0 & (i) \\
3A & +B &= 1 & (ii).
\end{array} \right.$$
From $(i)$ we see that $A=-B$. Plugging this into $(ii)$ yields $-3B+B=1$ which implies $B = -\dfrac{1}{2}$. Hence $A= \dfrac{1}{2}$. Therefore we have
$$\dfrac{1}{(2n+1)(2n+3)} = \dfrac{1}{4n+2} - \dfrac{1}{4n+6}.$$
This appears like it will be a telescoping series. So begin looking at partial sums: (recall a partial sum $S_N$ is the sum from $0$ to $N$ of the summand):
$$S_0 = \dfrac{1}{2} - \dfrac{1}{6},$$
$$S_1 = \left( \dfrac{1}{2} - \dfrac{1}{6} \right) + \left( \dfrac{1}{6} - \dfrac{1}{10} \right),$$
$$S_2 = \left( \dfrac{1}{2} - \dfrac{1}{6} \right) + \left( \dfrac{1}{6} - \dfrac{1}{10} \right) + \left( \dfrac{1}{10} - \dfrac{1}{14} \right)$$
$$S_3 = \left( \dfrac{1}{2} - \dfrac{1}{6} \right) + \left( \dfrac{1}{6} - \dfrac{1}{10} \right) + \left( \dfrac{1}{10} - \dfrac{1}{14} \right) + \left( \dfrac{1}{14} - \dfrac{1}{18} \right),$$
and so we observe
$$S_N= \left( \dfrac{1}{2} - \dfrac{1}{6} \right) + \left( \dfrac{1}{6} - \dfrac{1}{10} \right) + \left( \dfrac{1}{10} - \dfrac{1}{14} \right) + \left( \dfrac{1}{14} - \dfrac{1}{18} \right) + \ldots + \left( \dfrac{1}{4N+2} - \dfrac{1}{4N+6} \right) = \dfrac{1}{2} - \dfrac{1}{4N+6}.$$
Therefore we may compute
$$\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)(2n+3)} = \displaystyle\lim_{N \rightarrow \infty} S_N = \displaystyle\lim_{N \rightarrow \infty} \left( \dfrac{1}{2}-\dfrac{1}{4N+6} \right) = \dfrac{1}{2}.$$
Section 9.2 #42: Converge or not?
$$\displaystyle\sum_{n=0}^{\infty} \dfrac{3^n}{1000}.$$
Solution: This is a geometric series with $r=3$. Therefore it diverges.
Section 9.2 #43: Converge or not?
$$\displaystyle\sum_{n=1}^{\infty} \dfrac{n+10}{10n+1}.$$
Solution: Check the limit of the inside:
$$\displaystyle\lim_{n \rightarrow \infty} \dfrac{n+10}{10n+1} = \dfrac{1}{10} \neq 0.$$
Since the limit of the summand is not zero, we may conclude that the series diverges (Theorem 9.9, pg. 599).
Section 9.2 #48: Converge or not?
$$\displaystyle\sum_{n=0}^{\infty} \dfrac{3}{5^n}.$$
Solution: It converges because it is a geometric series with $r= \dfrac{1}{5}$.
Section 9.2 #64: Find all values of $x$ for which the series converges. For these values of $x$, write the sum of the series as a function of $x$:
$$\displaystyle\sum_{n=1}^{\infty} 5 \left( \dfrac{x-2}{3} \right)^n.$$
Solution: This series is geometric with $r=\dfrac{x-2}{3}$. For convergence, we require $-1 \lt r \lt 1$, so plug this expression in to get
$$-1 \lt \dfrac{x-2}{3} \lt 1.$$
Multiply by $3$ to get
$$-3 \lt x-2 \lt 3.$$
Add $2$ to both sides to get
$$-1 \lt x \lt 5.$$
Section 9.3 #30: Use the integral test to determine the convergence or divergence of the $p$-series:
$$\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{\frac{1}{2}}}.$$
Solution: Compute
$$\begin{array}{ll}
\displaystyle\int_1^{\infty} \dfrac{1}{x^{\frac{1}{2}}} \mathrm{d}x &= \displaystyle\int_1^{\infty} x^{-\frac{1}{2}} \mathrm{d}x \\
&= \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_1^b x^{-\frac{1}{2}} \mathrm{d}x \\
&= \dfrac{2}{3} \displaystyle\lim_{b \rightarrow \infty} x^{\frac{3}{2}} \Bigg|_1^b \\
&= \dfrac{2}{3} \displaystyle\lim_{b \rightarrow \infty} b^{\frac{3}{2}}-1 \\
&= \infty,
\end{array}$$
hence the integral diverges. Therefore the series also diverges.
Section 9.3 #38: Use the integral test to determine the convergence or divergence of the $p$-series:
$$\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{\pi}}.$$
Solution: Compute
$$\begin{array}{ll}
\displaystyle\int_1^{\infty} \dfrac{1}{x^{\pi}} \mathrm{d}x &= \dfrac{1}{-\pi+1} \displaystyle\lim_{b \rightarrow \infty} x^{-\pi+1} \Bigg|_1^b \\
&=\dfrac{1}{1-\pi} \displaystyle\lim_{b \rightarrow \infty} \dfrac{1}{b^{\pi-1}} - 1 \\
&= 0,
\end{array}$$
hence the integral converges. (Note: we can conclude convergence between $\pi-1 \approx 3.14-1 = 2.14$ and so the denominator is "growing large" as $b \rightarrow \infty$ while the numerator remains fixed.) Therefore the sum converges.
Section 9.3 #51: Find the positive values of $p$ for which the series converges:
$$\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{3}{p} \right)^n.$$
Solution: This series is a geometric series with ratio $r=\dfrac{3}{p}$. Geometric series convergee whenever $-1 \lt r \lt 1$ so plugging in the formula for $r$ yields $-1 \lt \dfrac{3}{p} \lt 1$. Since $p$ is in the denominator, it is useful to split this into two inequalities: $-1 \lt \dfrac{3}{p}$ and $\dfrac{3}{p} \lt 1$. In the first one, multiply by (the positive number) $p$ to get $-p \lt 3$ and then multiply by $-1$ to get $p \gt 3$. In the second one, multiply by $p$ to get $3 \lt p$. Notice that both of these mean the same thing: $p \gt 3$. (note: the series also converges when $p \lt -3$, but the problem asked for the positive values of $p$).
