# 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.
• Two series indeed can be combined: $\cos n + i \sin n = e^{in}$ – Sungjin Kim May 19 '13 at 2:27
• @i707107: yes, I figured there may be tricks from complex analysis, but my complex analysis is weak. – symplectomorphic May 19 '13 at 2:34
• Nonrigorously, it's "clear" that these are not absolutely convergent. On an almost arithmetic progression basis, $\left\lvert\cos(n)\right\rvert$ and $\left\lvert\sin(n)\right\rvert$ are close to $1$ (say, above $0.9$). So a subseries of these positive-termed series is "almost" harmonic. – alex.jordan Oct 29 '15 at 20:34

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 $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$.

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.