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Using the $p-$ test $\Rightarrow \sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ is convergent for $p>1$.

Using the comparison test $\Rightarrow$ $$a_n=\frac{\cos{nx}}{n^p},b_n=\frac{1}{n^p}$$

then $a_n$ is convergent if $p>1$ and divergent if $p\le 1$.

Is this the only necessary condition for convergence of $a_n$?

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  • $\begingroup$ Do you know anything about $x$? $\endgroup$
    – Hirshy
    Commented Jan 10, 2016 at 20:28
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    $\begingroup$ You can't use comparison test for sequences which aren't everywhere positive, so this doesn't work for $x$ not a multiple of $\pi$. Dirichlet test might be of help here. $\endgroup$
    – Wojowu
    Commented Jan 10, 2016 at 20:28
  • $\begingroup$ @Hirshy $x\in\mathbb{R}$. $\endgroup$
    – user300048
    Commented Jan 10, 2016 at 20:32
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    $\begingroup$ Then you should follow @Wojowu hint, as you don't neccesarily have strict positive $a_n$. $\endgroup$
    – Hirshy
    Commented Jan 10, 2016 at 20:34
  • $\begingroup$ Notice that when $p>1$, you can show that the series converges absolutely using the Comparison Test. $\endgroup$
    – user84413
    Commented Jan 10, 2016 at 20:36

1 Answer 1

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Since $$ \sum_{n=1}^\infty \left|\frac{\cos nx}{n^p} \right| \leq \sum_{n=1}^\infty \frac{1}{n^p}, $$ $\displaystyle \sum_{n=1}^\infty \frac{\cos nx}{n^p} $ is absolutely convergent for $p>1$, and hence convergent for $p>1.$

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