# Does $\sum_{n=1}^{\infty}(-1)^{n}\cos(\frac{\pi}{n})$ converges?

My attempt:

I used the fact that if $$\lim_{n \rightarrow \infty} a_{n} \neq 0 \Rightarrow \sum_{n=1}^{\infty} a_{n}$$ doesn´t converges. So, I checked that

\begin{align*} \lim_{n \rightarrow \infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right |=1 \end{align*} i.e., \begin{align*} \lim_{n \rightarrow \infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right |\neq0 \Longrightarrow \sum_{n=1}^{\infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right | \text{does not converges} \end{align*}

My doubt:

Is this implication correct? \begin{align} \sum_{n=1}^{\infty} \left |(-1)^{n}\cos(\frac{\pi}{n}) \right | \text{does not converges} \Longrightarrow \sum_{n=1}^{\infty} (-1)^{n}\cos(\frac{\pi}{n}) \text{ does not converges} \end{align}

• The last implication is false (e.g. $\sum (-1)^n/n$) but you don't need it since $|a_n|\not\to 0\implies a_n\not\to 0$.
– zwim
Feb 12, 2021 at 17:30
• That direct implication is not correct, but there was no need to take the absolute value. A limit that does not exist certainly does not equal $0$. Feb 12, 2021 at 17:30
• $\lim |a_n|=1$ means $\lim a_n\neq 0.$ Feb 12, 2021 at 18:19

Since $$\lim\limits_{n\to +\infty}{\left(-1\right)^{n}\cos{\left(\frac{\pi}{n}\right)}}$$ is not $$0$$, then your series diverges.