Does the series $\sum_{n=1}^\infty (-1)^n \frac{\cos(\ln(n))}{n^{\epsilon}},\,\epsilon>0$ converge?

Question

Numerical results showed that the series

$$Q=\sum_{n=1}^{\infty} (-1)^n \frac{\cos(\ln(n))}{n^{\epsilon}},\epsilon>0\tag{1}$$

with $\epsilon=10^{-6}$ converged to $-0.53259554096828...$

We can rewrite this series as:

$$Q=\sum_{k=0}^{\infty} a_k\tag{2}$$

$$a_k=\frac{\cos(\ln(2k+1))}{(2k+1)^{\epsilon}}-\frac{\cos(\ln(2k+2))}{(2k+2)^{\epsilon}}\tag{3}$$

Does this series converge?

Using a method in a related question, we can show that (with $m=2k+1$)

$$-a_k=m^{-1-\epsilon}\left(\epsilon\cos(\ln m)+\sin(\ln m)\right)+O(m^{-2-\epsilon}).\qquad m\to\infty \tag{4}$$

Thus

$$|a_k|\le m^{-1-\epsilon}\left(\epsilon|\cos(\ln m)|+|\sin(\ln m)|\right)+O(m^{-2-\epsilon})=O(m^{-1-\epsilon})=O(k^{-1-\epsilon})\tag{5}$$

Therefore the series in (1) is convergent.

Am I right?

Thanks- mike

Answer 1

Let $$ \eta(s)=\sum_{n=1}^\infty (-1)^{n+1}n^{-s}, $$ with $s=\sigma+it$. The real part of this series is $$ -\sum_{n=1}^\infty (-1)^n n^{-\sigma} \cos(t\ln(n)). $$ With $t=0$, the series converges (conditionally) for $\sigma>0$ by the alternating series test. A standard theorem about Dirichlet series then says the series converges (conditionally) for complex $s$ with $\text{Re}(s)>0$. If I recall correctly, the convergence is uniform in sectors $|\arg(s)|<\pi/2-\epsilon$.

It's worth noting that $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$.

Answer 2

Let me provide a more elementary answer.

As observed in the OP, we can rewrite the series as $$ Q=\sum_{k=0}^{\infty} a_k, $$ where $$ a_k=\frac{\cos(\ln(2k+1))}{(2k+1)^{\epsilon}}-\frac{\cos(\ln(2k+2))}{(2k+2)^{\epsilon}}. $$ Now $$ a_k=\frac{(2k+2)^{\epsilon}\cos(\ln(2k+1))-(2k+1)^{\epsilon}\cos(\ln(2k+2))}{\big((2k+1)(2k+2)\big)^{\epsilon}}\\=\frac{(2k+2)^{\epsilon}\big(\cos(\ln(2k+1))-\cos(\ln(2k+2))\big)}{\big((2k+1)(2k+2)\big)^{\epsilon}} + \frac{\big((2k+2)^{\epsilon}-(2k+1)^{\epsilon}\big)\cos(\ln(2k+2))}{\big((2k+1)(2k+2)\big)^{\epsilon}} \\ = b_k+c_k. $$ Now Mean Value Theorem for $f(x)=\cos(\ln x)$, with $f'(x)=-\sin(\ln x)/x$, provides that $$ \cos(\ln(2k+1))-\cos(\ln(2k+2))=\frac{\sin(\ln (2k+1+\xi_k))}{2k+1+\xi_k}, $$ for some $\xi_k\in(0,1)$, and thus $$ \lvert b_k\rvert\color{red}{=} \frac{\lvert\sin(\ln (2k+1+\xi_k))\rvert}{(2k+1+\xi_k)(2k+1)^\epsilon}\le \frac{1}{(2k+1)^{1+\epsilon}}, $$ which implies that $\sum_{k=0}^\infty \lvert b_k\rvert<\infty$. Meanwhile, applying the Mean Value Theorem once again to $g(x)=x^\epsilon$, we obtain that $$ \lvert c_k\rvert \le \frac{(2k+2)^{\epsilon}-(2k+1)^{\epsilon}}{\big((2k+1)(2k+2)\big)^{\epsilon}}\color{red}{=}\frac{\epsilon(2k+1+\xi_k)^{\epsilon-1}}{\big((2k+1)(2k+2)\big)^{\epsilon}}, \quad \xi_k\in(0,1), $$ and hence $$ \lvert c_k\rvert \le \frac{\epsilon(2k+2)^{\epsilon-1}}{\big((2k+1)(2k+2)\big)^{\epsilon}} =\frac{\epsilon}{(2k+1)^{\epsilon}(2k+2)}\le\frac{\epsilon}{(2k+1)^{1+\epsilon}}, $$ which also implies that $\sum_{k=0}^\infty \lvert c_k\rvert<\infty$.