Homework 13 Solution

FILIPE MOURA

11.2.1

a

$$\sum_{k=0}^{5} 2^k =1 + 2 + 4 +8 + 16 + 32 =63$$

$$\frac{1-2^6}{1-2}=2^6-1 = 63$$

b

$$\sum_{k=0}^{5}{\left(\frac{1}{100}\right)^k} =1 + \frac{1}{10... ...00000}+ \frac{1}{100000000000} =\frac{10101010101}{10000000000}$$

$$\frac{1-0.01^6}{1-0.001}=\frac{1-\frac{1}{1000000000000}}{1-\... ...ac{111111111111}{110000000000}= \frac{10101010101}{10000000000}$$

(notice that $11 \times 10101010101 = 101010101010 + 10101010101 = 111111111111$).

c

If we apply the formula for r=1, we get $\frac{0}{0}$.

In this case,

$$\sum_{k=0}^{n}{a 1^k} = (n+1)a$$

d

$$S_n - r S_n = \sum_{k=0}^{n}{a r^k} - \sum_{k=0}^{n}{a r^{k+1... ...- \sum_{k=1}^{n}{a r^k} - a r^{n+1} = a \left(1-r^{n+1} \right)$$

e

Using the formula from d,

$$S_n - r S_n = (1-r) S_n =a \left(1-r^{n+1} \right)$$

from where we get

$$S_n =a \left( \frac{1-r^{n+1}}{1-r} \right)$$

f

$3+6+ \cdots + 3072 = 3(1+2+4+8+ \cdots +1024)= 3\sum_{k=0}^{10}{2^k}=3 \left( \frac{1-2^{11}}{1-2} \right) = 3 \left(2^{11} -1 \right)= 6141$

11.2.2

a

$a_1 =1; a_2=\frac{1}{2}; a_5=\frac{1}{120}; a_{10}=\frac{1}{3628800}$

b

$S_1=1; S_2=\frac{5}{2}; S_5=\frac{163}{60}; S_{10}=\frac{9864101}{3628800}$

11.2.4

a

$a_1 =\frac{1}{5}; a_2=\frac{1}{25}; a_5=\frac{1}{3125}; a_{10}=\frac{1}{9765625}$

b

$S_1 =\frac{1-\left(\frac{1}{5}\right)^{2}}{1-\frac{1}{5}}=\frac{6}{5}; S_2=\fra... ...frac{1-\left(\frac{1}{5}\right)^{11}}{1-\frac{1}{5}}=\frac{12207 031}{97656255}$

c

S_n is increasing because $S_{n+1}-S_n=\left(\frac{1}{5}\right)^{n+1}>0$.

S_n is bounded above: $S_n=\frac{1-\left(\frac{1}{5}\right)^{n+1}}{1-\frac{1}{5}}=\frac{5- \left(\frac{1}{5}\right)^{n}}{4} < \frac{5}{4}$.

Therefore, S_n must converge.

d

$$\lim_{k \rightarrow \infty} S_n=\lim_{k \rightarrow \infty} \frac{5- \left(\frac{1}{5}\right)^{n}}{4} = \frac{5}{4}$$

e

$R_n=\sum_{k=n+1}^{\infty} \left(\frac{1}{5}\right)^{k} = \frac{1}{4} \left(\frac{1}{5}\right)^{n}$

$R_1 =\frac{1}{20}; R_2=\frac{1}{100}; R_5=\frac{1}{12500}; R_{10}=\frac{1}{39062500}$

f

Clearly, R_n>0.

$R_{n+1}-R_n=-\left(\frac{1}{5}\right)^{n+1}<0$

g

$$\lim_{k \rightarrow \infty} R_n=\frac{1}{4} \lim_{k \rightarrow \infty} \left(\frac{1}{5}\right)^{n} = 0$$

11.2.18

We have S_2n=1, S_2n+1=0. This implies that $\lim_{k \rightarrow \infty} S_n=\sum_{k=0}^{\infty} (-1)^k$ does not exist.

11.2.56

After the first bounce, the ball rebounds to a heigth of $4 \times \frac{2}{3}$ feet. The ball then falls from this heigth and rebounds to $4 \times \left(\frac{2}{3}\right)^2$ feet, etc.

Therefore, the total distance the ball travels is

$$4+ 2\times 4 \times \sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n =4 +\frac{8}{1- \frac{2}{3}} -8= 20$$

11.2.68

a

If the interval $\left[1, n+1 \right]$ is divided into n equal subintervals, and the left endpoint of the j^th interval is frac1j. Therefore, $L_n= \sum_{k=1}^{n} \frac{1}{k} =H_n$.

b

Since $\frac{1}{x}$ is a decreasing function on the interval $\left[1, \infty \right)$, L_n > I_N, for all $n \geq 1$. Since $I_n = \int^{n+1}_{1} \frac{d x}{x} =\ln (n+1)$, $\lim_{k \rightarrow \infty} I_n=\infty$ and $\lim_{k \rightarrow \infty} L_n=\lim_{k \rightarrow \infty} H_n=\sum_{k=1}^{\infty} \frac{1}{k}=\infty$.

11.3.1

a

For $k \geq 0$, $k+ 2^k \geq 2^k$. This implies $a_k \leq \frac{1}{2^k}$. Since

$\sum_{k=0}^{\infty} \frac{1}{2^k}$ converges, so does $\sum_{k=0}^{\infty} a_k$

b

$R_{10}=\sum_{k=11}^{\infty} a_k < \sum_{k=11}^{\infty} \frac{1}{2^k} = \frac{1}{2^{11}} \sum_{k=0}^{\infty} \frac{1}{2^k} = \frac{1}{2^{10}}$

c

Since $R_{10}<\frac{1}{2^{10}} \approx 0.00097656$, S_n has the desired accuracy if $n \geq 10$. A good estimate for the limit is

$$\sum_{k=0}^{10} a_k =\frac{127807216183}{75344540040}$$

d

Since a_k >0, $\sum_{k=0}^{10} a_k < \sum_{k=0}^{\infty} a_k$.

11.3.12

We have, for k>3, $\ln k > \ln 3$, or $\frac{1}{\ln k} < \frac{1}{\ln 3} <1$. Therefore, $\sum_{k=1}^{\infty} \frac{1}{{\ln k}^k}$ converges, since $\sum_{k=1}^{\infty} \frac{1}{(\ln 3)^k}$ converges to $\frac{1}{1-\frac{1}{\ln 3}} =\frac{\ln 3}{\ln 3 - 1}$.

11.3.14

$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \rightarrow \infty} \frac{(k+1)^2}{k^2} \frac{1}{k+1} =0 <1$$

11.3.15

$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \rightarrow \infty} \frac{2}{k+1} =0 <1$$

11.3.16

$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \rightarrow \infty} \frac{(k+1)^2}{k^2} \frac{1}{2} =\frac{1}{2} <1$$

11.3.21

We have

$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \ri... ...imes (2k+3)} = \lim_{k \rightarrow \infty} \frac{1}{2k+3} =0 <1$$

Therefore, by the ratio test, the series converges.

11.3.27

a

We have

$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \ri... ...{3}\right)^{k+1}}{1+\left(\frac{2}{3}\right)^{k}}= \frac{1}{2}$$

b

Since $\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} <1$, the series converges.

c

$$\sum_{k=1}^{\infty}{\left(\frac{1}{2^k} +\frac{1}{3^k} \right... ...c{1}{1-\frac{1}{2}} + \frac{1}{1-\frac{1}{3}} - 2 = \frac{3}{2}$$

(sum of two absolutely convergent series).