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(a) Find at least ten partial sums of the series. (Round your answers to five decimal places.) n=1cos(9n).\displaystyle\sum_{n=1}^\infty \cos(9n).

nnsns_n
1?\boxed{?}
2?\boxed{?}
3?\boxed{?}
4?\boxed{?}
5?\boxed{?}
6?\boxed{?}
7?\boxed{?}
8?\boxed{?}
9?\boxed{?}
10?\boxed{?}

(b) Does it appear that the series is convergent or divergent? If it is convergent, find the sum. (If the quantity diverges, enter DIVERGES.)

?\boxed{?}

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Question :

(a) find at least ten partial sums of the series. (round your answers to five decimal places.) n=1cos(9n).\displaystyle\sum_{n=1}^\infty \cos(9n).

nnsns_n
1?\boxed{?}
2?\boxed{?}
3?\boxed{?}
4?\boxed{?}
5?\boxed{?}
6?\boxed{?}
7?\boxed{?}
8?\boxed{?}
9?\boxed{?}
10?\boxed{?}

(b) does it appear that the series is convergent or divergent? if it is convergent, find the sum. (if the quantity diverges, enter diverges.)

?\boxed{?}

(a) find at least ten partial sums of the series. (round your answers to fiv | Doubtlet.com

Solution:

Neetesh Kumar

Neetesh Kumar | December 27, 2024

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This is the solution to Math 1c
Assignment: 11.2 Question Number 4
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Step-by-step solution:

Step 1: Analyze the series:

The series involves the general term:

an=cos(9n).a_n = \cos(9n).

The partial sum SnS_n is defined as:

Sn=k=1ncos(9k).S_n = \displaystyle\sum_{k=1}^n \cos(9k).

To evaluate the behavior of SnS_n, we need to compute the first 1010 partial sums.

Step 2: Compute the first 1010 partial sums:

Using a calculator or computational tool, evaluate:

  1. For n=1n = 1:

    S1=cos(91)=cos(9)0.911130.S_1 = \cos(9 \cdot 1) = \cos(9) \approx -0.911130.

  2. For n=2n = 2:

    S2=S1+cos(92)=0.911130+cos(18)0.91113+0.6603167=0.250813.S_2 = S_1 + \cos(9 \cdot 2) = -0.911130 + \cos(18) \approx -0.91113 + 0.6603167= -0.250813.

  3. For n=3n = 3:

    S3=S2+cos(93)=0.250813+cos(27)0.2508130.29213880=0.5429518.S_3 = S_2 + \cos(9 \cdot 3) = -0.250813 + \cos(27) \approx -0.250813 - 0.29213880 = -0.5429518.

  4. For n=4n = 4:

    S4=S3+cos(94)=0.5429518+cos(36)0.54295180.12796=0.67091.S_4 = S_3 + \cos(9 \cdot 4) = -0.5429518 + \cos(36) \approx -0.5429518 - 0.12796 = -0.67091.

  5. For n=5n = 5:

    S5=S4+cos(95)=0.67091+cos(45)0.67091+0.52532=0.14559.S_5 = S_4 + \cos(9 \cdot 5) = -0.67091 + \cos(45) \approx -0.67091 + 0.52532 = -0.14559.

  6. For n=6n = 6:

    S6=S5+cos(96)=0.14559+cos(54)0.145590.82930=0.9749.S_6 = S_5 + \cos(9 \cdot 6) = -0.14559 + \cos(54) \approx -0.14559 - 0.82930 = -0.9749.

  7. For n=7n = 7:

    S7=S6+cos(97)=0.9749+cos(63)0.9749+0.98589=0.01099.S_7 = S_6 + \cos(9 \cdot 7) = -0.9749 + \cos(63) \approx -0.9749 + 0.98589 = 0.01099.

  8. For n=8n = 8:

    S8=S7+cos(98)=0.01099+cos(72)0.010990.967250=0.95625.S_8 = S_7 + \cos(9 \cdot 8) = 0.01099 + \cos(72) \approx 0.01099 - 0.967250 = -0.95625.

  9. For n=9n = 9:

    S9=S8+cos(99)=0.95625+cos(81)0.95625+0.776685=0.17957.S_9 = S_8 + \cos(9 \cdot 9) = -0.95625 + \cos(81) \approx -0.95625 + 0.776685 = -0.17957.

  10. For n=10n = 10:

    S10=S9+cos(910)=0.17957+cos(90)0.179570.44807=0.62764.S_{10} = S_9 + \cos(9 \cdot 10) = -0.17957 + \cos(90) \approx -0.17957 - 0.44807 = -0.62764.

Step 3: Populate the table:

nnSnS_n
10.911130-0.911130
20.250813-0.250813
30.542952-0.542952
40.67091-0.67091
50.14559-0.14559
60.9749-0.9749
70.010990.01099
80.95625-0.95625
90.17957-0.17957
100.62764-0.62764

Step 4: Convergence or divergence:

The partial sums SnS_n do not settle to a fixed value as nn increases. Instead, the values oscillate due to the periodic nature of the cosine function. Therefore, the series diverges.

Final Answer:

(a)

nnsns_n
10.911130\boxed{-0.911130}
20.250813\boxed{-0.250813}
30.542952\boxed{-0.542952}
40.67091\boxed{-0.67091}
50.14559\boxed{-0.14559}
60.9749\boxed{-0.9749}
70.01099\boxed{0.01099}
80.95625\boxed{-0.95625}
90.17957\boxed{-0.17957}
100.62764\boxed{-0.62764}

(b)

  • The series diverges\boxed{\text{diverges}}

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