Exam 3 Review for series and summation PDF

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Summary

Review for exam 3. Great examples for series and limits. A lot of summation problems that are very helpful and wonderful practice!...

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Exam 3 Review 1) Assume the pattern continues and find a formula for the general term for the sequence 2 −1 2 , ,...} 2 5 1 2 6 24 120

a. {2, −1, , 3

b. {− , , − 2 3

4

, ,− 5

6

,

720 7

,...}

2) List the first 4 terms of the recursive series 1

a. Let 𝑎1 = 3 𝑎𝑛𝑑 𝑎𝑛+1 = (2𝑎𝑛 + 1) 3

b. Let 𝑎1 = −2 𝑎𝑛𝑑 𝑎𝑛+1 = (7 −

𝑎𝑛

5

).

3) If lim 𝑎𝑛 = 5 𝑎𝑛𝑑 lim 𝑏𝑛 = √13 𝑤𝑖𝑡ℎ 𝑏𝑛 ≠ 0 ∀𝑛. Then what is lim 𝑛→∞

𝑛→∞

4) Consider the series ∑ ∞ 𝑛=1 𝑎𝑛 . If 𝑎1 = 21 𝑎𝑛𝑑

𝑎𝑛+1 𝑎𝑛

=

1

5

𝑎𝑛

𝑛→∞ 𝑏𝑛

?

for all integers n greater than or

equal to n, then determine if the series is convergent or divergent (Hint: First write this out as a series, then solve) 5) Determine if the sequence diverges or converges, if so evaluate a. 𝑎𝑛 =

arctan (𝑛) 𝑛 2 𝑛

b. 𝑎𝑛 = (1 + )

𝑛 4𝑛 3 −𝑛+5

c. 𝑎𝑛 =

d. e.

, furthermore, determine if the series converges or diverges

2𝑛 3 +6𝑛2 −11 4𝑛 3 −𝑛+5 ∑∞ 𝑛=1 2𝑛 3 +6𝑛2−11 1 𝑎𝑛 = (−1)𝑛 𝑛2 (ln 𝑛 )2 𝑎𝑛 = √𝑛 5 𝑛

f. ∑ ∞ 𝑛=2 4 (− 6)

1 𝑛

g. ∑ ∞ 𝑛=3 3 (− 6) h. ∑ ∞ 𝑛=0

3𝑛+1

(−2)𝑛

 as a quotient of integers 6) Write 0.234678 7) Determine if the series converges or diverges a. ∑ ∞ 𝑛=1 ln ( b. ∑ ∞ 𝑛=1

𝑛

𝑛+1

)

3

√𝑛

√𝑛 3 +4𝑛+3

3 √ c. ∑ ∞ 𝑛=1 ( 3√𝑥 ) cos2 𝑛

d. ∑ ∞ 𝑛=1 𝑛 3 +𝑛 e. ∑ ∞ 𝑛=1 ( f.

1

𝑛 2 +5𝑛+6 𝑛−1

∑∞ 𝑛=1(−1)

) arctan 𝑛

g. ∑ ∞ 𝑛=1 ( h. i.

𝑛 2 +1 𝑛

)

2𝑛 2 +1 ∞ ln 𝑛 ∑ 𝑛=1 𝑛2 √2𝑛 2 +2 ∑∞ 𝑛=1 3𝑛+1 7𝑛

j. ∑ ∞ 𝑛=1 6𝑛 −1 (2𝑛!)

k. ∑ ∞ 𝑛=1 (𝑛!)2 l. ∑ ∞ 𝑛=1

𝑛+sin2 𝑛 𝑛 3 +1 1

m. ∑ ∞ 𝑛=1 tan ( 𝑛 ) 3

−1

√𝑛 n. ∑ ∞ 𝑛=1 𝑛(√𝑛+1) 1

o. ∑ ∞ 𝑛=1 (𝑛 3 + p. ∑ ∞ 𝑛=1

q. ∑ ∞ 𝑛=1

1

1 ) 3𝑛

𝑛 3976 (−1)𝑛−1 (ln 𝑛)𝑛 1

r. ∑ ∞ 𝑛=1 (ln ( 2))

𝑛

8) Assume ∑ 𝑎𝑛 converges to 𝑒 2 and ∑ 𝑏𝑛 converges to

1 3

a. What does ∑ 𝑎𝑛 𝑏𝑛 converge to? b. What does ∑ 𝑎𝑛 − 𝑏𝑛 converge to? c. What does ∑ 𝜋𝑏𝑛 converge to? 9) Approximate the sum of the series to 4 decimal points 𝑛 −2𝑛 , first determine how many terms this will take a. ∑∞ 𝑛=1(−1) 𝑛𝑒 10) For what value of p does the series converge a. ∑ ∞ 𝑛=1

1

𝑛 3𝑝 −𝑛2

11) Find the radius of convergence and interval of convergence of the series a. ∑ ∞ 𝑛=1

(−1)𝑛 4𝑛 √𝑛

𝑥𝑛

Answers 1) 2

a. 𝑎𝑛 = (−1) 𝑛+1 𝑛 b. 𝑎𝑛 = (−1)𝑛

2)

𝑛!

𝑛+1 7

a. 𝑎1 = 3, 𝑎2 = , 𝑎3 = 3

3) 4)

5

b. 𝑎1 = −2, 𝑎2 =

37 5

17 9

, 𝑎3 =

, 𝑎4 = 212 25

43

27

, 𝑎4 =

1087 125

√13 21 4

5) a. b. c. d. e. f. g.

0 𝑒2 2, series diverges by divergence test 0 0 50

33 −1 84

h. Diverges 6)

19537 83250

7) a. b. c. d. e. f. g. h. i. j. k. l. m. n.

Diverges (telescoping series) Converges (comparison test) Diverges (divergence test) Converges (comparison test) Converges to 1/3, telescoping series Diverges (AST) Converges absolutely (root test), all terms positive so converges Converges Limit comparison test and sheet at end of 11.1 Diverges (divergence test) Diverges (root test) Diverges (ratio test) Converges (comparison test) Diverge (limit comparison) Convergent (comparison)

o. Converges (p-series test and root test, converges absolutely and all terms positive so series converges) p. Converges, p series test q. Converges (AST) −ln (2)

r. Converges to 1+ln (2) 8) a.

𝑒2

3 2

b. 𝑒 − c.

𝜋 3

1

3

9) 5 𝑡𝑒𝑟𝑚𝑠, −0.1050 10) Type equation here.

1

1

1

11) Radius of convergence 4, interval of convergence (− 4 , ] 4...