# Convergence of a series involving ln() terms in the denominator of a fraction

• Amaelle
In summary, this person's approach to convergence is incorrect and they need to find a better approximation for the denominator.
Amaelle
Homework Statement
the convergence of a series (look at the image)
Relevant Equations
asymptotic comparison
good day
I want to study the convergence of this serie and want to check my approch

I want to procede by asymptotic comparison
artgln n ≈pi/2
n+n ln^2 n ≈n ln^2 n
and we know that
1/(n ln^2 n ) converge so the initial serie converge

many thanks in advance!

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This approach is far too hand wavy. You have no control about the approximations, and they even occur in the denominator! If you want to use the majority criterion you need an estimation
$$\dfrac{1}{n(1+\log^2(n))\tan^{-1}(\log(n))}\leq \dfrac{1}{n^c} \Longleftrightarrow n^c\leq n(1+\log^2(n))\tan^{-1}(\log(n))$$
for some ##c > 1##. This means you have to find a lower bound for the denominator which is still big enough.

docnet and Amaelle
Amaelle said:
I want to procede by asymptotic comparison
artgln n ≈pi/2
n+n ln^2 n ≈n ln^2 n
You should make this stronger. n+n ln^2 n > n ln^2 n > 0, so |1/(n + n ln^2 n )| < |1/(n ln^2 n )|.
Amaelle said:
and we know that
1/(n ln^2 n ) converge so the initial serie converge
Do you know that it is absolutely convergent? If so, say that.

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docnet and Amaelle
thank you so much so my approach , (if I write it as you did ) is correct? I can use the approximation lim artgln n ≈pi/2 when n tends to infinity?

fresh_42 said:
This approach is far too hand wavy. You have no control about the approximations, and they even occur in the denominator! If you want to use the majority criterion you need an estimation
$$\dfrac{1}{n(1+\log^2(n))\tan^{-1}(\log(n))}\leq \dfrac{1}{n^c} \Longleftrightarrow n^c\leq n(1+\log^2(n))\tan^{-1}(\log(n))$$
for some ##c > 1##. This means you have to find a lower bound for the denominator which is still big enough.
thank you very much , I got your point!

What is artg?

the function arctangent

Got it. It's certainly valid to say things like if n is large enough, artg(ln(n)) > 1, and then use that as a lower bound for the denominator.

Amaelle and FactChecker
Office_Shredder said:
Got it. It's certainly valid to say things like if n is large enough, artg(ln(n)) > 1, and then use that as a lower bound for the denominator.
thanks a lot!

## 1. What is the definition of convergence in a series involving ln() terms in the denominator of a fraction?

The convergence of a series involving ln() terms in the denominator of a fraction means that the series approaches a finite limit as the number of terms in the series increases.

## 2. How do you determine if a series involving ln() terms in the denominator of a fraction is convergent or divergent?

To determine if a series involving ln() terms in the denominator of a fraction is convergent or divergent, you can use the ratio test or the comparison test. If the limit of the ratio of consecutive terms is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent.

## 3. Can a series involving ln() terms in the denominator of a fraction converge to a negative value?

No, a series involving ln() terms in the denominator of a fraction cannot converge to a negative value. The terms in the series must be positive for the series to converge.

## 4. Is there a specific method for finding the sum of a series involving ln() terms in the denominator of a fraction?

Yes, there are specific methods for finding the sum of a series involving ln() terms in the denominator of a fraction. One method is to use the partial fraction decomposition technique to simplify the series and then use known summation formulas for ln() terms.

## 5. Can a series involving ln() terms in the denominator of a fraction converge if the ln() terms are raised to a power?

Yes, a series involving ln() terms in the denominator of a fraction can still converge if the ln() terms are raised to a power. However, the convergence or divergence of the series may depend on the value of the power and the other terms in the series.

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