conditionally convergence of $\{\cos(n)/\sqrt{n}\}$

Question

The problem is basically the title: prove that $\sum_{n=1}^\infty \frac{\cos(n)}{\sqrt{n}}$ is conditionally convergent.

I proved the convergence of the sum by Dirichlet's test, but i couldn't prove that $\sum_{n=1}^\infty \frac{|\cos(n)|}{\sqrt{n}}=\infty$.

Do you have any hint?

Answer 1

For a very easy proof of divergence, note that since $0 \leqslant |\cos n| \leqslant 1$, we have $$|\cos n| \geqslant \cos^2 n = \frac{1}{2} + \frac{ \cos 2n}{2}.$$

Hence,

$$\sum_{n=1}^m \frac{|\cos n|}{\sqrt{n}} \geqslant \sum_{n=1}^m \frac{1}{2\sqrt{n}} + \sum_{n=1}^m \frac{\cos 2n}{2\sqrt{n}}. $$

The second series on the right-hand side converges by the Dirichlet test, but the first series diverges by comparison with the divergent harmonic series $\frac{1}{2}\sum_{n \geqslant 1} \frac{1}{n} = \infty.$

Therefore, the series $\sum_{n \geqslant 1} \frac{|\cos n|}{\sqrt{n}}$ diverges.

Answer 2

Disproving the absolute convergence is a bit tricky.
An opportunity is given by the Fourier cosine series of $\left|\cos(x)\right|$:

$$\left|\cos(m)\right| = \frac{2}{\pi}+\frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{4n^2-1}\,\cos(nm) $$ where the series part $\frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{4n^2-1}\,\cos(nm)$ is absolutely convergent and with bounded partial sums (with respect to $m$). By invoking Dirichlet's test again it follows that the convergence of $\sum_{m\geq 1}\frac{\left|\cos(m)\right|}{\sqrt{m}}$ would imply the convergence of $\sum_{m\geq 1}\frac{2}{\pi\sqrt{m}}$, which is blatantly divergent: $$ \sum_{m=1}^{M}\frac{2}{\pi\sqrt{m}} = \frac{4}{\pi}\sqrt{M}+O(1) $$ follows by creative telescoping: $\frac{1}{\sqrt{m}}$ is well-approximated by the telescopic term $2\sqrt{m+\frac{1}{2}}-2\sqrt{m-\frac{1}{2}}$.

Answer 3

Here is a simple argument for divergence of $\sum_{n=1}^{\infty} \frac{|\cos(n)|}{1/\sqrt n}$.
Since cos(x) is zero only at multiplicities of $\frac{2k+1}{2}\pi$ and since these zeros are $\pi$ apart one of the three consecutive numbers 3k, 3k+1, 3k + 2 must be "far" from such zero. Say, $|cos(n)| >= c> 0$ for such n. Hence $$\sum_{n=1}^{\infty} \frac{|\cos(n)|}{1/\sqrt n} = $$ $$\sum_{k=1}^{\infty} ( \frac{|\cos(3k)|}{1/\sqrt 3k} + \frac{|\cos(3k + 1)|}{1/\sqrt (3k +1)} + \frac{|\cos(3k+2)|}{1/\sqrt (3k+2)} ) > = $$ $$\sum_{k=1}^{\infty} ( \frac{c}{1/\sqrt (3k+2)}$$ and since the series $\sum \frac{1}{1/\sqrt (3k+2)}$ diverges the original is divergent as well.