Methods for determining the convergence of $\sum\frac{\cos n}{n}$ or $\sum\frac{\sin n}{n}$
As far as I know, the textbook approach to determining the convergence of series like $$\sum_{n=1}^\infty\frac{\cos n}{n}$$ and $$\sum_{n=1}^\infty\frac{\sin n}{n}$$ uses Dirichlet's test, which involves bounding the partial sums of the cosine or sine terms. I have two questions:
- Are there any other approaches to seeing that these series are convergent? I'm mostly just interested to see what other kinds of arguments might be made.
- What's the best way to show that these two series are only conditionally convergent? I don't even know the textbook approach to that question.
2 Answers
$\begingroup$Hint for 2)
$$\sum_{n=1}^{\infty} \frac{|\cos n|}{n} \geq \sum_{n=1}^{\infty} \frac{\cos^2 n}{n}=\sum_{n=1}^{\infty} \frac{1+\cos {2n}}{2n}$$
Convergence of $\sum_{n=1}^{\infty}\frac{\cos{2n}}{2n}$, and divergence of $\sum_{n=1}^{\infty}\frac{1}{2n}$ gives the divergence.
The same method applies to $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$.
$\endgroup$ $\begingroup$Thanks for the nice question, below is a sketch for part 1, which may be a few inches from a rigorous proof.
Let us consider the complex sum $$ S \equiv \sum_{n = 1}^\infty \frac{ e^{in} }{n}, $$ The real and imaginary sums are the desired cosine and sine sums, as pointed out by i707107. Then $$ e^{-i} S = \sum_{n = 1}^\infty \frac{ e^{i(n-1)} }{n} = 1 + \sum_{n = 1}^\infty \frac{ e^{i n} }{n + 1}, $$ and $$ (1 - e^{-i}) \, S = -1 + \sum_{n = 1}^\infty \left(\frac{1}{n} - \frac{1}{n + 1} \right) e^{i n} = -1 + \sum_{n = 1}^\infty \frac{e^{i n}}{n \, (n + 1)}. $$
So $$ \begin{align} |(1 - e^{-i}) \, S| &\le 1 + \sum_{n = 1}^\infty \left| \frac{e^{i n}}{n \, (n + 1)} \right| \\ &= 1 + \sum_{n = 1}^\infty \frac{1}{n \, (n + 1)} \\ &= 1 + \sum_{n = 1}^\infty \left(\frac{1}{n} - \frac{1}{n + 1}\right) = 2. \end{align} $$ This means $|S|$ is finite, and so are the real and imaginary parts.
$\endgroup$More in general
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