The three series Σ An Σ Bn, and Σ Cn have terms 1 n7 2. EMBIMBiM8 3. Σ An Use the Limit Comparison Test to compare the following series to any of the above series. For each of the series below, you must enter two letters. The first is the letter (A,B or C) of the series above that it can be legally compared to with the Limit Comparison Test. The second is C if the given series converges, or D if it diverges. So for instance, if you believe the series converges and can be compared with series C above, you would enter CC; or if you believe it diverges and can be compared with series A, you would enter AD. 6n² +8n6 2n7 +10n³ - 2 8n³+n²-8n 10n 10 6n⁹ +3 Bn = 2n²+ n² 561n⁹ +10n² +6 1 n² Cn= 1 n
The three series Σ An Σ Bn, and Σ Cn have terms 1 n7 2. EMBIMBiM8 3. Σ An Use the Limit Comparison Test to compare the following series to any of the above series. For each of the series below, you must enter two letters. The first is the letter (A,B or C) of the series above that it can be legally compared to with the Limit Comparison Test. The second is C if the given series converges, or D if it diverges. So for instance, if you believe the series converges and can be compared with series C above, you would enter CC; or if you believe it diverges and can be compared with series A, you would enter AD. 6n² +8n6 2n7 +10n³ - 2 8n³+n²-8n 10n 10 6n⁹ +3 Bn = 2n²+ n² 561n⁹ +10n² +6 1 n² Cn= 1 n
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:The three series Σ Ani Σ Bn, and Σ Cn have terms
1
n7
1.
2.
3.
∞
n=1
∞
n=1
∞
An
Use the Limit Comparison Test to compare the following series to any of the above
series. For each of the series below, you must enter two letters. The first is the letter (A,B,
or C) of the series above that it can be legally compared to with the Limit Comparison
Test. The second is C if the given series converges, or D if it diverges. So for instance, if
you believe the series converges and can be compared with series C above, you would
enter CC; or if you believe it diverges and can be compared with series A, you would
enter AD.
n=1
=
B₁
6n2 + 8n6
2n7+10n³2
8n³ + n² - 8n
10n 10 6n⁹ +3
2n² + n²
561n⁹ +10n² + 6
1
n²
Cn=
=
1/20
n
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