Explain how the Limit Comparison Test works. THEOREM 10.15 Limit Comparison TestLet Ea, and Eb, be series with positive terms and letlimL.1. If 0 < L < ∞ (that is, L is a finite positive number), then Ea, and Eb,either both converge or both diverge.2. If L = 0 and Eb, converges, then Ea, converges.3. If L = ∞ and Eb, diverges, then Ea, diverges.

Let ∑ak and ∑bk be two positive term series, and limk→∞akbk=L. 1. If 0<L<∞(i.e. L is a finite…

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Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

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Explain how the Limit Comparison Test works.

THEOREM 10.15 Limit Comparison Test
Let Ea, and Eb, be series with positive terms and let
lim
L.
1. If 0 < L < ∞ (that is, L is a finite positive number), then Ea, and Eb,
either both converge or both diverge.
2. If L = 0 and Eb, converges, then Ea, converges.
3. If L = ∞ and Eb, diverges, then Ea, diverges.

Expert Solution

Step 1

Let $\sum a_{k}$ and $\sum b_{k}$ be two positive term series, and $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L$ .

1. If $0 < L < \infty$ (i.e. L is a finite positive number) Then $\sum a_{k}$ and $\sum b_{k}$ both converge or diverge together.

2. If $L = 0$ and $\sum b_{k}$ converges, then $\sum a_{k}$ also converges.

3. $L = \infty$ and $\sum b_{k}$ diverges, then $\sum a_{k}$ also diverges.

Step 2

1. Since $a_{k} > 0$ , $b_{k} > 0$

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L > 0$

But $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L \neq 0$

Now $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L$

$\Rightarrow$ given $\in > 0$ , $\exists$ a positive integer m such that

$|\frac{a_{k}}{b_{k}} - L| < \in \forall k > m$

$\Rightarrow L - \in < \frac{a_{k}}{b_{k}} < L + \in \forall k > m$

Choose $\in > 0$ such that $L - \in > 0$ .

Let $L - \in = H, L + \in = K$ where $H, K > 0$

$∴ H b_{k} < a_{k} < K b_{k} \forall k > m . . . (*)$

Case-1: When $\sum a_{k}$ is convergent

From $(*)$ $H b_{k} < a_{k} \forall k > m$

$\Rightarrow b_{k} < \frac{1}{H} a_{k} \forall k > m (∴ H > 0)$

Since $\sum a_{k}$ is convergent, $\sum b_{k}$ is also convergent.

Case.II: When $\sum a_{k}$ is divergent.

From $(*)$ , $a_{k} < K b_{k} \forall k > m$

$\Rightarrow \frac{a_{k}}{K} < b_{k} \forall k > m (∵ K > 0)$

Since $\sum a_{k}$ is divergent, $\sum b_{k}$ is also divergent.

Case.III: When $\sum b_{k}$ is convergent.

From (*), $a_{k} < K b_{k} \forall k > m$

Since $\sum b_{k}$ is convergent, $\sum a_{k}$ is also convergent.

Case:IV: When $\sum b_{k}$ is divergent.

From (*), $H b_{k} < a_{k} \forall k > m$

Since $\sum b_{k}$ is divergent, $\sum a_{k}$ is also divergent.

Hence $\sum a_{k}$ and $\sum b_{k}$ converge or diverge together.

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