TITLE: Does the series $\sum_{n=1}^\infty (-1)^n \frac{\cos(\ln(n))}{\sqrt{n}}$ converge?
QUESTION [9 upvotes]: Numerical results for $m=1$ to $2000$ showed that the series
$$Q(m)=\sum_{n=1}^m (-1)^n \frac{\cos(\ln(n))}{\sqrt{n}}$$
converged to $-0.63986...$
Does the series 
$$\sum_{n=1}^{\infty} (-1)^n \frac{\cos(\ln(n))}{\sqrt{n}}$$
converge?
here is plot of $Q(m)$ vs. $m$

REPLY [11 votes]: Convergence
The Euler Maclaurin Sum Formula says that, for some constants $C$ and $S$,
$$
\sum_{k=1}^n\frac{\cos(\log(k))}{\sqrt{k}}
=\sqrt{n}\left[\frac25\cos(\log(n))+\frac45\sin(\log(n))\right]+C+O\left(n^{-1/2}\right)\tag{1}
$$
and
$$
\sum_{k=1}^n\frac{\sin(\log(k))}{\sqrt{k}}
=\sqrt{n}\left[\frac25\sin(\log(n))-\frac45\cos(\log(n))\right]+S+O\left(n^{-1/2}\right)\tag{2}
$$
Writing the alternating sum as the difference of twice the even terms minus all the terms, we get
$$
\begin{align}
&\sum_{k=1}^{2n}(-1)^k\frac{\cos(\log(k))}{\sqrt{k}}\\
&=2\sum_{k=1}^n\frac{\cos(\log(2k))}{\sqrt{2k}}
-\sum_{k=1}^{2n}\frac{\cos(\log(k))}{\sqrt{k}}\\
&=\sqrt2\sum_{k=1}^n\frac{\cos(\log(2)+\log(k))}{\sqrt{k}}
-\sum_{k=1}^{2n}\frac{\cos(\log(k))}{\sqrt{k}}\\
&=\sqrt2\cos(\log(2))\sum_{k=1}^n\frac{\cos(\log(k))}{\sqrt{k}}\\
&-\sqrt2\sin(\log(2))\sum_{k=1}^n\frac{\sin(\log(k))}{\sqrt{k}}
-\sum_{k=1}^{2n}\frac{\cos(\log(k))}{\sqrt{k}}\\
&=\sqrt2\cos(\log(2))\sqrt{n}\left[\frac25\cos(\log(n))+\frac45\sin(\log(n))\right]\\
&-\sqrt2\sin(\log(2))\sqrt{n}\left[\frac25\sin(\log(n))-\frac45\cos(\log(n))\right]\\
&-\sqrt{2n}\left[\frac25\cos(\log(2n))+\frac45\sin(\log(2n))\right]\\
&+\left[\vphantom{\frac25}\sqrt2\cos(\log(2))-1\right]C-\sqrt2\sin(\log(2))\,S+O\left(n^{-1/2}\right)\\[6pt]
&=\left[\vphantom{\frac25}\sqrt2\cos(\log(2))-1\right]C-\sqrt2\sin(\log(2))\,S+O\left(n^{-1/2}\right)\tag{3}
\end{align}
$$
Thus, the series converges to
$$
\left[\vphantom{\frac25}\sqrt2\cos(\log(2))-1\right]C-\sqrt2\sin(\log(2))\,S\tag{4}
$$

Computing the Sum
Using the Euler-Maclaurin Sum Formula to compute $C$ and $S$, we get
$$
C=0.1439364270771890603243896664837216\tag{5}
$$
and
$$
S=0.7220997435316730891261751345803249\tag{6}
$$
Therefore, using $(4)$, we get
$$
\sum_{k=1}^\infty(-1)^k\frac{\cos(\log(k))}{\sqrt{k}}=-0.6398619139367474311364313137759324\tag{7}
$$

The Asymptotic Expansions
Including more terms of the Euler-Maclaurin Sum Formula, we have
$$
\sum_{k=1}^n\frac{\cos(\log(k))}{\sqrt{k}}
=\sqrt{n}\left[a_s(n)\cos(\log(n))+a_c(n)\sin(\log(n))\vphantom{\tfrac25}\right]+C\tag{8}
$$
and
$$
\sum_{k=1}^n\frac{\sin(\log(k))}{\sqrt{k}}
=\sqrt{n}\left[a_s(n)\sin(\log(n))-a_c(n)\cos(\log(n))\vphantom{\tfrac25}\right]+S\tag{9}
$$
where
$$
\begin{align}
\hspace{-1cm}\small a_s(n)\,&\small=\frac25+\frac1{2n}-\frac1{24n^2}-\frac7{1920n^4}+\frac{491}{193536n^6}-\frac{11903}{4423680n^8}+\frac{822169}{181665792n^{10}}+O\left(\frac1{n^{12}}\right)\tag{10}\\[4pt]
\hspace{-1cm}\small a_c(n)\,&\small=\frac45-\frac1{12n^2}+\frac{19}{2880n^4}-\frac{157}{96768n^6}+\frac{10039}{15482880n^8}+\frac{146483}{2452488192n^{10}}+O\left(\frac1{n^{12}}\right)\tag{11}
\end{align}
$$
$(8)-(11)$ were used with $n=1000$ to compute $(5)$ and $(6)$ to over $34$ places of precision.
Note that $(1)$ and $(2)$ are just truncated versions of $(8)-(11)$.

Zeta Function
As achille hui has noted in a comment, we have
$$
\begin{align}
\sum_{k=1}^\infty(-1)^k\frac{\cos(\log(k))}{\sqrt{k}}
&=\mathrm{Re}\left[\sum_{k=1}^\infty(-1)^ke^{i\log(k)}k^{-1/2}\right]\\
&=\mathrm{Re}\left[\sum_{k=1}^\infty(-1)^kk^{-1/2+i}\right]\\
&=-\mathrm{Re}\left[\eta\left(1/2-i\right)\vphantom{\tfrac12}\right]\\[6pt]
&=\mathrm{Re}\left[\left(2^{1/2+i}-1\right)\zeta\left(1/2-i\right)\right]
\end{align}
$$
This is probably why riemann-zeta appears in the tags for this question.
In this answer, it is shown that the series for $\eta(s)$ converges for $\mathrm{Re}(s)\gt0$. So that answer offers another method to show that the sum in this question converges.