# Does $\sum_{n=2}^{\infty}\frac{\cos\left(\frac{\pi n^2}{n+1}\right)}{(\log n)^2}$ converge?

I'm pretty sure the sum converges by Abel or Dirichelet, I just have no idea how to tackle the numerator. Any tips would be appreciated!

• Something like this $\;\cos\left(\pi\frac{n^2}{n+1}\right)/\log(n)^2 = (-1)^{n-1}\cos\left(\frac{\pi}{n+1}\right)/\log(n)^2$? Commented Jan 25, 2014 at 20:55
• Hawk, you edited it incorrectly. Check the original post before you edit. The parentheses were placed like I have them. Commented Jan 25, 2014 at 20:57
• @DavidMitra $\cos\left(\pi\frac{n^2}{n+1}\right) = \cos\left(\pi\frac{n^2-1 + 1}{n+1}\right) = \cos\left(\pi(n-1) + \frac{\pi}{n+1}\right)$ Commented Jan 25, 2014 at 21:07
• @achillehui Right. Sorry... Commented Jan 25, 2014 at 21:08

## 1 Answer

Note that $$\frac{n^2}{n+1}= \frac{(n^2 + n) - (n+1) + 1}{n+1} = n-1+\frac{1}{n+1}$$ From there, we note that \begin{align} \cos\left(\frac{\pi n^2}{n+1}\right) &= \cos\left(\pi(n-1) + \frac{\pi}{n+1}\right) \\&= \cos\left(\pi(n-1)\right)\cos\left(\frac{\pi}{n+1}\right) - \sin\left(\pi(n-1)\right)\sin\left(\frac{\pi}{n+1}\right) \\&= (-1)^n\cos\left(\frac{\pi}{n+1}\right) \end{align} Noting that for all $n \geq 2$, $0 < \frac{\pi}{n+1}< \pi/2$ so that $\cos(\frac{\pi}{n+1})$ is strictly positive and less than $1$. It follows that $$0<\frac{\cos(\frac{\pi}{n+1})}{(\log(n))^2} < \frac{1}{(\log(n))^2}$$ Now, in order to apply the alternating series test, we would need to show that $\frac{\cos(\frac{\pi}{n+1})}{(\log(n))^2}$ is decreasing for sufficiently large $n$.

In order to do so, note that $$\frac{d}{dx} \frac{\cos(\frac{\pi}{x+1})}{(\log(x))^2} = \frac{\pi \sin(\frac{\pi}{x+1})}{(1+x^2)(\log(x))^2} -\frac{2 \cos(\frac{\pi}{x+1})}{x(\log(x))^3}$$ If we may find an $x_0$ so that the above is negative for all $x>x_0$, then we will have shown that the sequence eventually decreases, as is required.

As David Mitra points out in the comment below, $x_0 = 2$ works perfectly well.

• Perhaps some justification that $\cos(\pi/(n+1))/\log^2 n$ is decreasing should be made. It's not immediate to see, at least. Commented Jan 25, 2014 at 21:21
• @DavidMitra I think you're right. I'm trying to think of a slick justification that this should be the case. Commented Jan 25, 2014 at 21:37
• I'm not very slick, so I just computed the derivative. It's negative when $n>2$ and $${\pi\log n\over (n+1)^2} \sin{\pi\over n+1}-{2\over n}\cos{\pi\over n+1}$$ is negative. This is the case for sufficiently large $n$. Commented Jan 25, 2014 at 21:45