2
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The familiar formula for the Riemann zeta function:

$$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}\right) \mbox{ is true for } \Re(s)>1$$

adding one more term of the Euler-Maclaurin formula we get:

$$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right) \mbox{ which appears to be true for }\Re(s)>0$$

adding yet one more term of the Euler-Maclaurin formula we get:

$$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)} -\frac{k^{-s}}{2} \right) \mbox{ which appears to be true for } \Re(s)>-1$$

From that on in general, it appears that:

$$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left(\sum\limits_{n=1}^{n=k} \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}+\sum\limits_{r=1}^{q-1} \frac{B_{2 r} k^{-2 r-s+1} \left(\prod _{i=0}^{2 r-2} (i+s)\right)}{(2 r)!}\right)$$

is true whenever: $\Re(s)>-(2q-1)$ where $q=1,2,3,4,5,...$

Is this last generalization a simple fact of the Euler Maclaurin formula for the analytic continuation of the Riemann zeta function?

Clear[n, k, s];
Limit[Sum[1/n^s, {n, 1, k}], k -> Infinity, Assumptions -> Re[s] > 1]
Limit[Sum[1/n^s, {n, 1, k}] + 1/k^(s - 1)/(s - 1), k -> Infinity, 
 Assumptions -> Re[s] > 0]
Limit[Sum[1/n^s, {n, 1, k}] + 1/k^(s - 1)/(s - 1) - (k^(-s))/2, 
 k -> Infinity, Assumptions -> Re[s] > -1]
q = 9;
Limit[Sum[1/n^s, {n, 1, k}] + k^(1 - s)/(s - 1) - (k^(-s))/2 + 
  Sum[BernoulliB[2*r]/((2*r)!)*Product[s + i, {i, 0, 2*r - 2}]*
    k^(-s - 2*r + 1), {r, 1, q - 1}], k -> Infinity, 
 Assumptions -> Re[s] > -(2*q - 1)]
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  • $\begingroup$ This answer math.stackexchange.com/a/47183/8530 appears to answer it in the positive, at least in a similar manner. $\endgroup$ – Mats Granvik Mar 23 at 20:59
  • 1
    $\begingroup$ If this is "a simple fact", why nobody tell that? $\endgroup$ – Aleksey Druggist Mar 25 at 9:51
  • $\begingroup$ Perhaps this is relevant to the subject: "The Prime Number Theorem" by J.G.O. Jameson, Cambridge University Press, 2003, p.112 $\endgroup$ – Aleksey Druggist Mar 27 at 8:12
  • $\begingroup$ @AlekseyDruggist Semiclassical in the chat room sent me the following link: math.ucdavis.edu/~tracy/courses/math205A/… where it says at the end: "By repeating the above argument we see that we have analytically continued the Riemann zeta-function to the right-half plane σ > 1 − k, for all k = 1, 2, 3, . . .." which answers my question when you index the Bernoulli numbers differently. $\endgroup$ – Mats Granvik Mar 27 at 15:18

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