I got this wrong on a quiz and it would be really helpful if someone could tell me how to solve this.
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The only difficulty with direction comparison is that it requires you to establish an inequality. I'd opt for a limit comparison test against the convergent p-series Σ1/n². Note that
lim ln(1 + 1/n²)/(1/n²) = 1.
n->∞
As this limit is positive and finite, both series converge or both series diverge. Since Σ1/n² is known to be convergent, you can conclude that your series is also convergent.
As your series consists of all non-negative terms, the convergence is absolute.
lim ln(1 + 1/n²)/(1/n²) = 1.
n->∞
As this limit is positive and finite, both series converge or both series diverge. Since Σ1/n² is known to be convergent, you can conclude that your series is also convergent.
As your series consists of all non-negative terms, the convergence is absolute.
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I made a typo on the limits. It should have been n²/(n² + 1), best answer was selected during the time I was editing.
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You were supposed to show that the absolute value of the terms converges which then implies the original series also converges.
On n ∈ [1, ∞), ln(1 + 1/n²) > 0, so we only need to show that ∑ ln(1 + 1/n²) converges.
Let an = ln(1 + 1/n²) and bn = 1/n²
Then we have
lim (n→∞) ln(1 + 1/n²) / 1/n² = lim (n→∞) -2n / (n² + 1) = 1 > 0
Note that ∑ 1/n² is a convergent series (in fact, Euler showed that it evaluates to π²/6!), and therefore by the Limit Comparison Test, ∑ ln(1 + 1/n²) must also converge.
Yin
On n ∈ [1, ∞), ln(1 + 1/n²) > 0, so we only need to show that ∑ ln(1 + 1/n²) converges.
Let an = ln(1 + 1/n²) and bn = 1/n²
Then we have
lim (n→∞) ln(1 + 1/n²) / 1/n² = lim (n→∞) -2n / (n² + 1) = 1 > 0
Note that ∑ 1/n² is a convergent series (in fact, Euler showed that it evaluates to π²/6!), and therefore by the Limit Comparison Test, ∑ ln(1 + 1/n²) must also converge.
Yin