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Math  /  Calculus

QuestionUse the limit comparison test to determine whether ∑n=5∞an=∑n=5∞7n+35n6+2n2+5\sum_{n=5}^{\infty} a_{n}=\sum_{n=5}^{\infty} \frac{7 n+3}{5 n^{6}+2 n^{2}+5} converges or diverges. (a) Choose a series ∑n=5∞bn\sum_{n=5}^{\infty} b_{n} with terms of the form bn=1npb_{n}=\frac{1}{n^{p}} and apply the limit comparison test. Write your answer as a fully reduced fraction. For n≥5n \geq 5, lim⁡n→∞anbn=lim⁡n→∞\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} □\square □\square □\square (b) Evaluate the limit in the previous part. Enter ∞\infty as infinity and −∞-\infty as -infinity. If the limit does not exist, enter DNE. □\square lim⁡n→∞anbn=□\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\square (c) By the limit comparison test, does the series converge, diverge, or is the test inconclusive? Choose □\square

Studdy Solution

STEP 1

1. We are using the limit comparison test to determine the convergence or divergence of the series ∑n=5∞an=∑n=5∞7n+35n6+2n2+5\sum_{n=5}^{\infty} a_{n} = \sum_{n=5}^{\infty} \frac{7n+3}{5n^{6}+2n^{2}+5}.
2. We need to choose a comparison series ∑n=5∞bn\sum_{n=5}^{\infty} b_{n} with terms of the form bn=1npb_{n} = \frac{1}{n^{p}}.
3. The limit comparison test states that if lim⁡n→∞anbn=c\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = c, where 0<c<∞0 < c < \infty, then both series either converge or diverge together.

STEP 2

1. Choose an appropriate comparison series ∑n=5∞bn\sum_{n=5}^{\infty} b_{n}.
2. Calculate the limit lim⁡n→∞anbn\lim_{n \to \infty} \frac{a_{n}}{b_{n}}.
3. Evaluate the limit.
4. Determine the convergence or divergence of the series based on the limit comparison test.

STEP 3

Choose a series ∑n=5∞bn\sum_{n=5}^{\infty} b_{n} with terms of the form bn=1npb_{n} = \frac{1}{n^{p}}.
Since the dominant term in the denominator of an=7n+35n6+2n2+5a_{n} = \frac{7n+3}{5n^{6}+2n^{2}+5} is 5n65n^{6}, we choose bn=1n6b_{n} = \frac{1}{n^{6}}.

STEP 4

Calculate the limit:
lim⁡n→∞anbn=lim⁡n→∞7n+35n6+2n2+51n6\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \lim_{n \to \infty} \frac{\frac{7n+3}{5n^{6}+2n^{2}+5}}{\frac{1}{n^{6}}}
Simplify the expression:
=lim⁡n→∞(7n+3)⋅n65n6+2n2+5= \lim_{n \to \infty} \frac{(7n+3) \cdot n^{6}}{5n^{6}+2n^{2}+5}

STEP 5

Continue simplifying:
=lim⁡n→∞7n7+3n65n6+2n2+5= \lim_{n \to \infty} \frac{7n^{7} + 3n^{6}}{5n^{6} + 2n^{2} + 5}
Divide every term by n6n^{6}:
=lim⁡n→∞7n+35+2n4+5n6= \lim_{n \to \infty} \frac{7n + 3}{5 + \frac{2}{n^{4}} + \frac{5}{n^{6}}}
As n→∞n \to \infty, the terms 2n4\frac{2}{n^{4}} and 5n6\frac{5}{n^{6}} approach zero:
=lim⁡n→∞7n+35= \lim_{n \to \infty} \frac{7n + 3}{5}

STEP 6

Evaluate the limit:
=lim⁡n→∞7n+35=75⋅lim⁡n→∞n=∞= \lim_{n \to \infty} \frac{7n + 3}{5} = \frac{7}{5} \cdot \lim_{n \to \infty} n = \infty

STEP 7

By the limit comparison test, since the limit is ∞\infty, the test is inconclusive. However, since bn=1n6b_{n} = \frac{1}{n^{6}} converges (as a pp-series with p=6>1p=6 > 1), and the limit was not finite, we cannot conclude about the convergence of ∑an\sum a_{n} directly from this test.
The limit comparison test is inconclusive in this case. To determine the behavior of ∑an\sum a_{n}, further analysis or a different test would be needed.

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