(Limit Comparison Test) Suppose (xn) and (yn) are positive sequences such that lim (xn/yn) exists. (i) If lim (xn/yn) ≠ 0, the series ∑xn and ∑yn either both converge or they both diverge. (ii) If lim (xn/yn) = 0 and the series ∑yn converges, then ∑xn also converges.

Question

(Limit Comparison Test)

Suppose [latex](x_{n})[/latex] and [latex](y_{n})[/latex] are positive sequences such that lim [latex](x_{n}/y_{n})[/latex] exists.

(i) If lim [latex](x_{n}/y_{n}) ≠ 0,[/latex] the series [latex]\sum{x_{n}}[/latex] and [latex]\sum{y_{n}}[/latex] either both converge or they both diverge.

(ii) If lim [latex](x_{n}/y_{n}) = 0[/latex] and the series [latex]\sum{y_{n}}[/latex] converges, then [latex]\sum{x_{n}}[/latex] also converges.

Accepted Answer

In case (i) we know that lim [latex]x_{n}/y_{n}[/latex] must be a positive number. According to Theorems 3.2 and 3.3, the sequence [latex](x_{n}/y_{n})[/latex] is bounded away from 0, that is, there are two positive constants M and K and a positive integer N such that

n ≥ N ⇒ [latex]M ≤ \frac{x_{n}}{y_{n}} ≤ K[/latex]

⇒ [latex]My_{n} ≤ x_{n} ≤ Ky_{n}.[/latex]

Using the comparison test, we now conclude that the convergence of one of the two series implies the convergence of the other.

In case (ii) there is a positive integer N such that

n ≥ N ⇒ [latex]0 ≤ \frac{x_{n}}{y_{n}} ≤ 1[/latex]

⇒ [latex]0 ≤ x_{n} ≤ y_{n},[/latex]

so, again by the comparison test, [latex]\sum{x_{n}}[/latex] converges if [latex]\sum{y_{n}}[/latex] converges.

Question 4.T.7: (Limit Comparison Test) Suppose (xn) and (yn) are positive s...

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