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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

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    $\begingroup$ Yes. (And I am not sure the present question should not be considered a duplicate of the previous one.) $\endgroup$ – Did Sep 19 '14 at 9:11
  • $\begingroup$ @Did. Thanks again for the conformation. This result is probably obvious to you. But it might not be obvious to new comers like me. I was hesitating to include this part in the original question or in a new question. alex.jordan commented in the previous question that if $\epsilon>1/2$, it is quite easy to prove the convergence. It is a pleasant surprise to me that this method also works for $\epsilon>0$. Next time I will definitely add such thing in the original question. And please feel free to mark it as duplicate. $\endgroup$ – mike Sep 19 '14 at 9:29
  • $\begingroup$ @Jean-ClaudeArbaut Thanks. I corrected it. $\endgroup$ – mike Sep 19 '14 at 9:36
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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 $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$.

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  • $\begingroup$ I do not see how you applied the alternating series test. According to my sources, the alternating series test only applies to series of the form $\sum(-1)^na_n$ where $a_{n+1}<a_n$ for all $n$, but that is not the case since $\cos(t\ln(n))$ changes sign infinitely often. Also, I think you "standard theorem" is faulty, since the Riemann zeta is a Dirichlet series but diverges for $\Re(s)\in(-\infty,1]$. $\endgroup$ – Simply Beautiful Art Mar 15 '17 at 21:27
  • $\begingroup$ @SimplyBeautifulArt, you misunderstand. You apply the alternating series test when $t = 0$. This implies that the series converges conditionally for $s = \sigma + it$ with $\sigma > 0$ and $t = 0$. Then you use the standard theorem that says once you have conditional convergence at a point $\sigma_0$ (i.e. with $t = 0$), you also have conditional convergence for all $s$ in the right-half plane $\Re(s) > \sigma_0$. $\endgroup$ – Peter Humphries Mar 15 '17 at 21:35
  • $\begingroup$ The Dirichlet series representation of the Riemann zeta function does not converge conditionally at the point $s = \sigma + it$ with $t = 0$ for all $\sigma < 1$, so the standard theorem is not faulty. $\endgroup$ – Peter Humphries Mar 15 '17 at 21:37
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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 $\sum_{k=0}^\infty \lvert c_k\rvert<\infty$.

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  • $\begingroup$ Yiorgos: Thanks a lot for the answer! But I do not quite follow 2 specific steps with $\color{red}{=}$ signs. $\endgroup$ – mike Sep 20 '14 at 19:59
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    $\begingroup$ @mike: See edit. $\endgroup$ – Yiorgos S. Smyrlis Sep 20 '14 at 20:19
  • $\begingroup$ Yiorgos: Thank a lot! $\endgroup$ – mike Sep 20 '14 at 20:36

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