Is this infinite series convergent or divergent?

[Lotto]

Homework Statement

I want to find out if this infinite series is convergent or divergent: $\sum_{n=2}^{\infty} \frac{1}{\ln^2 n} \left|\cos{\frac{\pi n^2}{n+1}}\right|$

Relevant Equations

We need to use a convergence test. I would use either Abel's or Dirichlet's test.

I have no idea where to start. Could you help me to start somewhere? I suppose I have to rewrite the series somehow, but I don't know how so that it helps

 

[fresh_42]

$\dfrac{\pi n^2}{n+1}=\pi n \cdot \dfrac{1}{1+(1/n)}$ is close to $\pi n$ for sufficiently large $n.$ This means that the cosine of it is around $\pm 1$ and can probably be approximated by $1/2$ from below. Then we get a lower bound $\displaystyle{\sum_{n>1}\dfrac{1}{2\log^2(n)}}$ which diverges.

 

[Lotto]

$\dfrac{\pi n^2}{n+1}=\pi n \cdot \dfrac{1}{1+(1/n)}$ is close to $\pi n$ for sufficiently large $n.$ This means that the cosine of it is around $\pm 1$ and can probably be approximated by $1/2$ from below. Then we get a lower bound $\displaystyle{\sum_{n>1}\dfrac{1}{2\log^2(n)}}$ which diverges.

OK, I see, but here we have to prove that the cosine in an absolute value is always greater than $1/2$ for prime numbers greater than a certain $n_0$. Could we prove that the series diverges in a different way?

Btw, how do we know that the series of $\frac{1}{\ln^2 n}$ diverges? I can't find any test that shows it.

 

[fresh_42]

OK, I see, but here we have to prove that the cosine in an absolute value is always greater than $1/2$ for prime numbers greater than a certain $n_0$.

What do you mean by prime numbers? I see the problem. Let me think a bit.

Could we prove that the series diverges in a different way?

Different to what, and why? We need to rule out that the cosine factor doesn't get so small that it can ruin the divergence of $\dfrac{1}{\log^2(n)}.$ There is no way to avoid that step.

Btw, how do we know that the series of $\frac{1}{\ln^2 n}$ diverges? I can't find any test that shows it.

Such tests are ultimately always comparison tests. In this case, we need a divergent lower bound. The standard example is $\displaystyle{\sum \dfrac{1}{n}}.$ Can you show that $\dfrac{1}{2\log^2(n)}>\dfrac{1}{n}$?

 

[fresh_42]

I do not want to post the full solution, so here are my hints: calculate the distance between $\pi n$ and $\dfrac{\pi n^2}{n+1}$ and show that it is between $2/3 \pi$ and $\pi.$ Then determine the possible range for $\dfrac{\pi n^2}{n+1}$ and consider the cases where $n$ is even and $n$ is odd and take the cosines. Show that the absolute values of the possible cosines are at least $0.5\,.$ Finally compare $\dfrac{1}{2\log^2(n)}$ with $1/n.$

 

[Lotto]

I do not want to post the full solution, so here are my hints: calculate the distance between $\pi n$ and $\dfrac{\pi n^2}{n+1}$ and show that it is between $2/3 \pi$ and $\pi.$ Then determine the possible range for $\dfrac{\pi n^2}{n+1}$ and consider the cases where $n$ is even and $n$ is odd and take the cosines. Show that the absolute values of the possible cosines are at least $0.5\,.$ Finally compare $\dfrac{1}{2\log^2(n)}$ with $1/n.$

Well, I tried that a little bit differently. Here are my steps:

We can write $\left|\cos{\frac{\pi n^2}{n+1}}\right|$ as $\left|\cos{\frac{\pi}{n+1}}\right|$. If we take a derivative of a function $f(x) = \left|\cos{\frac{\pi}{x+1}}\right|$, we get $f'(x) = \tan{\left(\frac{\pi}{x+1}\right)} \cdot \frac{\pi}{(x+1)^2} \cdot \left|\cos{\frac{\pi}{x+1}}\right|$. This is always positive for all natural numbers, so our sequence is always increasing and approaches $1$ at infinity, so from a certain $n_0$, it has to be greater than $1/2$ for example.

And because from a certain $n'_0$ it is true that $n >2\ln^2 n$ (because both sequences are positivie and increasing and the limit of $\frac{n}{2 ln^2 n}$ is $\infty$) which is equivalent to $\frac{1}{2 \ln ^2 n} > \frac 1n$, our series diverges.

The solution might be too complicated, but is it correct?

 

[fresh_42]

Well, I tried that a little bit differently. Here are my steps:

We can write $\left|\cos{\frac{\pi n^2}{n+1}}\right|$ as $\left|\cos{\frac{\pi}{n+1}}\right|$. If we take a derivative of a function $f(x) = \left|\cos{\frac{\pi}{x+1}}\right|$, we get $f'(x) = \tan{\left(\frac{\pi}{x+1}\right)} \cdot \frac{\pi}{(x+1)^2} \cdot \left|\cos{\frac{\pi}{x+1}}\right|$. This is always positive for all natural numbers, so our sequence is always increasing and approaches $1$ at infinity, so from a certain $n_0$, it has to be greater than $1/2$ for example.

And because from a certain $n'_0$ it is true that $n >2\ln^2 n$ (because both sequences are positivie and increasing and the limit of $\frac{n}{2 ln^2 n}$ is $\infty$) which is equivalent to $\frac{1}{2 \ln ^2 n} > \frac 1n$, our series diverges.

The solution might be too complicated, but is it correct?

It looks good, but I would add why the first formula is true. It is not obvious to me without calculations, and the same is true for the derivative. Given that you made no mistake, it looks correct to me, and, by the way, very nice!

This is again an example of my usual motto when approaching a problem: "Remove what disturbs the most!" This was in this case the cosine term.

My solution was more basic:
\begin{align*}
\dfrac{2}{3}\pi<\left|\pi n - \dfrac{\pi n^2}{n+1}\right|&=\dfrac{\pi n}{n+1}= \dfrac{\pi}{1+\dfrac{1}{n}}<\pi \\
\pi (n-1)&\leq \dfrac{\pi n^2}{n+1}\leq \pi (n- (2/3))\\
n\text{ even }\, &: \, -1\leq \cos\left(\dfrac{\pi n^2}{n+1}\right)\leq -0.5\\
n\text{ odd }\, &: \, 1\geq \cos\left(\dfrac{\pi n^2}{n+1}\right)\geq 0.5\\
\left|\cos\left(\dfrac{\pi n^2}{n+1}\right)\right|&\geq \dfrac{1}{2}
\end{align*}