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

[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!

 

[fresh_42]

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.

 

[FactChecker]

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 )|.

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.

 

[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?

 

[Amaelle]

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!

 

[Office_Shredder]

What is artg?

 

[Amaelle]

the function arctangent

 

[Office_Shredder]

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]

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!