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Section 9.1 #29: Determine the convergence or divergence of the sequence with the given nth term. If it converges, find the limit:
an=5n+2.
Solution: The sequence converges. Compute
lim
Section 9.1 #36: Determine the convergence or divergence of the sequence with the given nth 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).