Calculus I, II, and III: A Problem-Based Approach with Early Transcendentals:

Section Solutions

1. Write the first five terms of the sequence in simplest form.

   1. $\dsp \frac{11}{3}, \frac{9}{3}, \frac{7}{3}, \frac{5}{3}, \dots$

   2. $\dsp 1, \frac{\sqrt{2}}{2}, 0, \frac{-\sqrt{2}}{2}, -1, \dots$

   3. $\dsp \frac{5}{2}, \frac{11}{4}, \frac{29}{8}, \dots$

2. Write a formula for each sequence. I.e. $x_n = \_\_\_$ for $n = 1,2,3,\dots\text{.}$

   1. $\dsp x_n = \frac{3n}{5^n} \cdots$

   2. $\dsp (-1)^n \cdots$

   3. $\dsp (-1)^{n+1} \frac{n}{n+1} \cdots$

3. What is the least upper bound and the greatest lower bound? Prove it.

   1. GLB = 0, LUB = 2

   2. GLB = 2/3 , LUB = 9/7

4. Is the sequence monotonic or not? Prove it.

   1. decreasing

   2. not monotonic

   3. increasing

5. Use limits to find whether or not the sequence converges.

   1. converges to 0

   2. converges to 1

   3. converges to 0

   4. converges to $e^{-1}$

6. N = 3001

7. N = the first positive integer larger than $\dsp \frac{3}{2\epsilon}$

1. List, in simplest form, the first three partial sums for the series and write a formula for the $N^{th}$ partial sum.

   1. $\dsp 1 , \frac{4}{3} , \frac{13}{9} , \frac{40}{27} , \dots, S_n= \frac{1-3^{n+1}}{1-3}\frac{1}{3^{n}} = \frac{1 - (\frac{1}{3})^{n+1}}{1-\frac{1}{3}}$

   2. $\dsp 3 , 3\frac{6}{5} , 3\frac{31}{25} , \dots, S_n = 3 \frac{1-5^{n+1}}{(1-5)5^{n}}= 3\frac{1 - (\frac{1}{5})^{n+1}}{1-\frac{1}{5}}$

2. What does the geometric series converge to?

   1. 1

   2. 9

3. $\dsp \sum_{n=3}^\infty \frac{5}{3n} = \frac{5}{3} \sum_{n=3}^\infty \frac{1}{n} > \sum_{n=3}^\infty \frac{1}{n}$ which diverges

4. Apply the integral test to determine if the series converges.

   1. converges

   2. diverges

5. Determine if these series converge via one of the $n^{th}$ term test, integral test, p-series test, comparison test, or limit comparison test.

   1. converges by limit comparison with $\dsp \frac{1}{n^2}$

   2. converges by limit comparison with $\dsp \frac{1}{n^2}$

   3. diverges, $n^{th}$ term test

   4. converges, geometric series

   5. diverges, $n^{th}$ term test

   6. separate the two series, one harmonic (diverges), one geometric (converges)

   7. separate the two series, both geometric (converges)

   8. separate the two series, one constant (diverges by $n^th$ term test), one converges by integral or p-series test

   9. limit comparison with $\dsp \frac{1}{n^2}$

   10. converges, geometric series

   11. limit comparison with $\dsp \frac{1}{n^3}$

   12. limit comparison with $\dsp \frac{1}{\sqrt{n}}$

   13. converges by limit comparison with $\dsp \frac{1}{n^2}$

   14. converges by comparison to $\dsp \frac{1}{n^2}$

   15. converges by limit comparison with $\dsp \frac{1}{n}$

   16. converges by comparison to $\dsp \frac{1}{n^2}$

   17. converges by comparison to $\dsp \frac{1}{n^2}$

   18. converges by limit comparison with $\dsp \frac{1}{3^n}$

   19. converges by comparison to $\dsp \frac{1}{n^2}$

   20. converges, integral test

6. Determine whether the alternating series is convergent or divergent.

   1. converges

   2. converges

   3. diverges, terms don't tend to zero

   4. diverges, rewrite it so that you can see that $a_n = (5/3)^n$

   5. converges

   6. converges

   7. diverges

   8. diverges

   9. converges

   10. converges

   11. diverges, terms don't tend to zero

   12. converges, rewrite it without a trig function

7. Conditionally Convergent, Absolutely Convergent, or Divergent?

   1. D

   2. CC

   3. AC, to get absolutely convergent, convert $\log_4$ to $\ln$ and use integral test

   4. AC

   5. D

   6. AC, ratio test + that clever natural log trick for limits

   7. D

   8. CC

   9. AC

   10. CC

   11. this is a neat problem

   12. this is a really neat problem

8. Intervals of convergence.

   1. $(-1,1)$

   2. $(-4,4)$

   3. $(-1/3,1/3)$

   4. $(2,8)$

   5. $(-1,1)$

   6. $(-1,1)$

   7. no interval, diverges for all $x \neq -3$

   8. no interval, diverges for all $x \neq 0$

   9. $(-8,0)$

   10. $(-1,1)$

   11. this is a neat problem

   12. $(1,3)$

9. Taylor Series (just the first few terms)

   1. $\dsp -1+2\, \left( x-1 \right) -3\, \left( x-1 \right) ^{2}+4\, \left( x-1 \right) ^{3}-5\, \left( x-1 \right) ^{4}+6\, \left( x-1 \right) ^{5}$

   2. $\dsp 2x -8x^3/3! +32 x^5/5! - 128 x^7/7! \dots$

   3. $\dsp 3+3\,\ln \left( 3 \right) \left( x-1 \right) + \frac{3}{2} \left( \ln \left( 3 \right) \right) ^{2} \left( x-1 \right) ^{2}+ \frac{1}{2} \left( \ln \left( 3 \right) \right) ^{3} \left( x-1 \right) ^{3}+\frac{1}{8} \left( \ln \left( 3 \right) \right) ^{4} \left( x-1 \right) ^{4}+$ $\dsp \frac{1}{40} \left( \ln \left( 3 \right) \right) ^{5} \left( x-1 \right) ^{5}$

   4. $\dsp {e^{2}}+{e^{2}} \left( x-2 \right) + \frac{1}{2}{e^{2}} \left( x-2 \right) ^ {2}+\frac{1}{6}{e^{2}} \left( x-2 \right) ^{3}+\frac{1}{24}{e^{2}} \left( x-2 \right) ^{4}+{\frac {1}{120}}\,{e^{2}} \left( x-2 \right) ^{5}$

   5. $\dsp -x+\frac{\pi}{2} +{\frac {1}{6}} \left( x-\frac{\pi}{2} \right) ^{3}-{\frac {1}{120}} \left( x-\frac{\pi}{2} \right) ^{5}$

   6. $\dsp x-{\frac {1}{2}}{x}^{2}+{\frac {1}{3}}{x}^{3}-{\frac {1}{4}}{x}^{4}+{ \frac {1}{5}}{x}^{5}$

10. Use multiplication, substitution, differentiation, and integration whenever possible to find a series representation of each of the following functions.

    1. $\dsp \sum_{n=0}^\infty (-1)^n \frac{x^{2(n+1)}}{(2n)!}$

    2. $\dsp \sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{(2n+1)!}$

    3. $\dsp 5\sum_{n=0}^\infty \frac{3^n}{n!}x^{n+1}$

    4. $\dsp 2 \sum_{n=0}^\infty (n-1)r^n$ You know the series for $\dsp \frac{1}{1-r}.$