The three series An, Bn, and Cn have terms An = 1 / n9, Bn = 1 / n5, Cn = 1 / n.

Question

The three series An, Bn, and Cn have terms An = 1 / n9, Bn = 1 / n5, Cn = 1 / n. 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. 8n5 + n2 - 8n / 10n14 - 8n11 + 5 5n5 + n9 / 35n14 + 10n5 + 8 8n2 + 8n8 / 5n9 + 10n3 - 5

Explanation / Answer

1) Tn = (8n^5+n^2-8n)/(10n^14-8n^11+5) = order(n^5)/order(n^14) = order(1/n^9)

so, lets choose, An = 1/n^9

we know that using term-test (http://en.wikipedia.org/wiki/Term_test), 1/n^p is convergent if (p>1), else divergent.

therefore, lim n->inf Tn/An = 4/5

Therefore, series is convergent according to term test.

so, AC

2) Tn = (n^9+5n^5)/(935n^14+10n^5+8) = order(n^9)/order(n^14) = order(1/n^5)

lets choose, Bn = 1/n^5

we know that using term-test (http://en.wikipedia.org/wiki/Term_test), 1/n^p is convergent if (p>1), else divergent.

therefore, lim n->inf Tn/An = 1/935

Therefore, series is convergent according to term test.

lets choose, Cn = 1/n

we know that using term-test (http://en.wikipedia.org/wiki/Term_test), 1/n^p is convergent if (p>1), else divergent.

therefore, lim n->inf Tn/An = 8/5

Therefore, series is divergent according to term test.

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