# If $\sum\limits_nz_n$ converges and $|\arg(z_n)|\leq\theta<\frac{\pi}{2}$ for every $n$ then $\sum\limits_nz_n$ converges absolutely [duplicate]

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Prove that if $\sum\limits_{n=1}^\infty z_n$ converges and there exists $\theta\in\mathbb{R}$ such that $|\arg(z_n)|\leq\theta<\frac{\pi}{2}$ for all $n\in\mathbb{Z^+}$, then it converges absolutely.

I tried invoking the convergence of $(z_n)$ to $0$, but couldn't find a second step or a route. Could someone give me some hints please?

Added later:

For all $n\in\mathbb{Z^+},z_n=\Re z_n+i\Im z_n$. Since $\sum\limits_{n=1}^\infty z_n$ converges $\sum\limits_{n=1}^\infty \Re z_n$ converges. But for all $n\in\mathbb{Z^+}$, $$\Re z_n=|z_n|\cos(\arg z_n)\geq|z_n|\cos\theta\geq0.$$ Now invoking the comparison test the convergence of $\sum\limits_{n=1}^\infty |z_n|\cos\theta$ is immediate. Hence $\sum\limits_{n=1}^\infty |z_n|$ converges, i.e. $\sum\limits_{n=1}^\infty z_n$ converges absolutely.

## marked as duplicate by Gabriel Romon, Lee David Chung Lin, RRL, Lord Shark the Unknown, CesareoFeb 2 at 7:58

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• Hint: Use $$\Re z_n\geqslant\cos\theta\cdot|z_n|$$ – Did Dec 11 '16 at 9:22
• @Did Is my answer added later alright? – Janitha357 Dec 11 '16 at 10:22
• Yes.    – Did Dec 11 '16 at 10:35