Convergence of a series: question

Question

I know
$$ \lim_{n \to \infty}\left\{\,% {2\cos\left({\left[k - 1\right]\pi \over n-3}\right) - 2\cos\left(k\pi \over n + 1\right)}\,\right\} = 0, $$ for $k \in \mathbb{N}$. $$ \mbox{Can I conclude}\quad \lim_{n \to \infty}\sum_{k = 1}^{n} \left\vert\,{2\cos\left({\left[k - 1\right]\pi \over n-3}\right) - 2\cos\left(k\pi \over n + 1\right)}\,\right\vert = 0\ {\large ?} $$ I think I can't. Thank you for your help.

Answer 1

As $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$$ we can't conclude that $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{k} = 0} $$

In your case $$2 \cos \left(\frac{\pi (k-1)}{n-3}\right)-2 \cos \left(\frac{\pi k}{n+1}\right)=\\-4 \sin \left(\frac{\pi (k-1)}{2 (n-3)}-\frac{\pi k}{2 (n+1)}\right) \sin \left(\frac{\pi (k-1)}{2 (n-3)}+\frac{\pi k}{2 (n+1)}\right)$$ and $$\left|\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n -4 \sin \left(\frac{\pi (k-1)}{2 (n-3)}-\frac{\pi k}{2 (n+1)}\right) \sin \left(\frac{\pi (k-1)}{2 (n-3)}+\frac{\pi k}{2 (n+1)}\right)\right)\right|=4$$

Answer 2

$$\lim_{n \to \infty}\left\{\,% {2\cos\left({\left(k - 1\right)\pi \over n-3}\right) - 2\cos\left(k\pi \over n + 1\right)}\,\right\} = 0$$ is true because for large values of $n$ $$2\cos\left({\left(k - 1\right)\pi \over n-3}\right) - 2\cos\left(k\pi \over n + 1\right)=\frac{(2k-1)\pi^2}{n^2}+O\left(\frac{1}{n^3}\right)$$ But, as an example $$\sum_{k=1}^n\frac{(2k-1)\pi^2}{n^2}=\pi^2$$ In fact, you have two summations of cosines where the arguments are in arithmetic progression, and, after simplifications, omitting the absolute value,

$$\sum_{k=1}^n \left\{\,% {2\cos\left({\left(k - 1\right)\pi \over n-3}\right) - 2\cos\left(k\pi \over n + 1\right)}\,\right\}=1+\cos \left(\frac{5 \pi (n-2)}{2 (n-3)}\right) \csc \left(\frac{\pi }{2 (n-3)}\right) \to -4$$

Quoting @Raffaele's comment "the sum of the abs values if greater or equal to the abs of the sum", using the absolute value of the sum. For the infinite sum, the result would be $ > 4$.

For example, using $n=15$, without absolute values, we should get $\left(-\sqrt{3}-\sqrt{2+\sqrt{3}}\right)\approx -3.66390$ while with absolute values the result would be $$4+\sqrt{3}+3 \sqrt{2+\sqrt{3}}-4 \cos \left(\frac{\pi }{8}\right)-4 \cos \left(\frac{3 \pi }{16}\right)\approx 4.50621$$