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How do we evaluate this summation:

$$H(b, c)=\sum_{n=1}^\infty \frac {\cos(c \log(n))} {n^b}$$ where $b, c$ are some positive constants such that $0<b<1$.

I know (if I am correct) that this series is convergent, but I can't find a method to calculate the sum. Can someone please comment?

Ricky
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stack.tarandeep
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    $\Re{;\zeta(b+c i)}$ should do the job! In fact providing a (better) regularized result since convergence of $H$ is questionable! – Raymond Manzoni Sep 30 '23 at 11:06
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    I don't think it converges. – user10354138 Sep 30 '23 at 11:09
  • 'syms b real syms c real syms n integer assume (b>0) assume (c>0) term = cos(c*log(n+1))/(n+1)^b; y = vpa(symsum(term, n, 1, inf)); Sumval0=symsum(term, n, 1, inf) if isinf(y) fprintf('divergent\n'); else fprintf('convergent\n'); disp(Sumval0); end' – stack.tarandeep Sep 30 '23 at 11:25
  • with the above code matlab says its convergent. what am i doing wrong here? @user10354138 can you please clarify? – stack.tarandeep Sep 30 '23 at 11:28
  • I don't use MATLAB. If MATLAB reports it as convergent then it is a bug in MATLAB and you should ask them for the explanation. – user10354138 Sep 30 '23 at 11:31
  • @Masacroso and raymond. MATLAB using the code above says the series is convergent. can you please comment on what is that is wrong i am doing here ? – stack.tarandeep Sep 30 '23 at 11:34
  • Matlab appears to tell that the sum is not infinite not that it is convergent (I don't use Matlab but evaluation of this kind of series using acceleration convergence will return a finite result but a result depending of precision and thus of no value. What value do you get for $H(0.5,0)$ ?). – Raymond Manzoni Sep 30 '23 at 11:39
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    To understand what is going on look at this answer : it shows the partial results of $\displaystyle\sum_{k=1}^N \frac 1{k^{b+ci}}$ for increasing values of $N$. See the growing disk at the end when $N\to +\infty$ (you want the real part of this series as $N\to +\infty$ => values oscillating always more at the end!) . – Raymond Manzoni Sep 30 '23 at 11:44

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