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

Question

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. $m$

Answer 1

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 $$ \left[\vphantom{\frac25}\sqrt2\cos(\log(2))-1\right]C-\sqrt2\sin(\log(2))\,S\tag{4} $$

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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 $$ \sum_{k=1}^\infty(-1)^k\frac{\cos(\log(k))}{\sqrt{k}}=-0.6398619139367474311364313137759324\tag{7} $$

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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 $(8)-(11)$.

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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.

Answer 2

We multiply the original series by $(-1)$ and obtain

$$-Q(m)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos(\ln(n))}{\sqrt{n}}=\sum_{k=0}^{\infty} a_k\tag{0}$$ with$$a_k=\frac{\cos(\ln(2k+1))}{\sqrt{2k+1}}-\frac{\cos(\ln(2k+2))}{\sqrt{2k+2}}\tag{1}$$

Set $2k+1=m$. Then, when $m\to\infty$,

$$\ln(m+1)=\ln m +m^{-1}+O(m^{-2}). \tag{2}$$

Substituting (2) into (1) and expanding the result as a series in $m^{-1/2}$ leads to

$$a_k=m^{-3/2}\left(\frac{1}{2}\cos(\ln m)+\sin(\ln m)\right)+O(m^{-5/2}).\qquad m\to\infty \tag{3}$$

So $$|a_k|\le m^{-3/2}\left(\frac{1}{2}|\cos(\ln m)|+|\sin(\ln m)|\right)+O(m^{-5/2})=O(m^{-3/2})=O(k^{-3/2})\tag{4}$$

Therefore the original series in (0) is convergent.