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Citing Wikipedia, the Riemann zeta function is the analytic continuation of

$$ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$

The series itself is only convergent in the right half complex plane where $Re(s)>1$.

What I would like to understand is how the series behaves near the line $1+it$, i.e. the boundary of this right half plane. On the boundary line we have $|\zeta(1+it)|<\infty$. Lets look at the specific example of $t=5$. for real $\epsilon>0$ I would write

$$ \zeta(1+5i) = \lim_{\epsilon\to 0}\sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon+5i}} $$

and I assume that in this order of taking the two limits, this is actually a correct equation, since otherwise $\zeta$ would not be an analytic continuation.

Does this mean that $\sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon+5i}}$ is bounded for each $\epsilon>0$? But if it is bounded yet does not converge for $\epsilon\to0$, does it mean that the sum, if understood as $\lim_{n\to\infty} \sum^n$, oscillates around the value of $\zeta(1+5i)$? What happens when $\epsilon$ goes towards $0$?

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