Absolute convergence of sum{n=1 to infinity} (ln(1+ (1/n^2)))
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Absolute convergence of sum{n=1 to infinity} (ln(1+ (1/n^2)))

[From: ] [author: ] [Date: 12-04-13] [Hit: ]
Euler showed that it evaluates to π²/6!), and therefore by the Limit Comparison Test, ∑ ln(1 + 1/n²) must also converge.......
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.

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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
1
keywords: infinity,ln,convergence,Absolute,to,sum,of,Absolute convergence of sum{n=1 to infinity} (ln(1+ (1/n^2)))
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