# A double sum and its relation to a simple sum, is this an identity for any complex number $S=a+i b$ and any integers n and t?

Does:

$$\sum _{m=1}^t \lim_{s\to \text{S}} \, \zeta (s) \sum _{k=(m-1) n+1}^{m n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

equal:

$$\lim_{s\to \text{S}} \, \zeta (s) \sum _{k=1}^{n t} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

for any complex number $S$ and any integers $n$ and $t$?

Which as a Mathematica program is:

S = 1.23456789 + I*9.87654321
Monitor[Table[
Chop[N[Table[
Sum[Limit[
Zeta[s]*Sum[(1 - If[Mod[k, n] == 0, n, 0])/
k^(s - 1), {k, (m - 1)*n + 1, m*n}], s -> S], {m, 1,
t}], {n, 1, 12}]] -
Table[
Limit[Zeta[s]*
Sum[(1 - If[Mod[k, n] == 0, n, 0])/k^(s - 1), {k, 1, t*n}],
s -> S], {n, 1, 12}]], {t, 1, 6}], t]


Some background information on the problem. Mathematica knows that logarithms can be calculated as:

$\log(n) = \lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$

Logarithms can be added to give the logarithm of the product of their argument,

$\lim_{s\to 1} \, \left(1-\frac{1}{6^{s-1}}\right) \zeta (s) = \lim_{s\to 1} \, \left(1-\frac{1}{2^{s-1}}\right) \zeta (s) + \lim_{s\to 1} \, \left(1-\frac{1}{3^{s-1}}\right) \zeta$

$\log(6) = \log(2) + \log(3)$

This works only when the limits lets $s \to 1$.

Trying the same for example with the limit letting $s \to 2$, it fails, as can be seen in this example:

$\lim_{s\to 2} \, \left(1-\frac{1}{6^{s-1}}\right) \zeta (s) = \lim_{s\to 2} \, \left(1-\frac{1}{2^{s-1}}\right) \zeta (s) + \lim_{s\to 2} \, \left(1-\frac{1}{3^{s-1}}\right) \zeta$

which gives us:

$\lim_{s\to 2} \, \left(1-\frac{1}{6^{s-1}}\right) \zeta (s) = \lim_{s\to 2} \, \left(1-\frac{1}{2^{s-1}}\right) \zeta (s) + \lim_{s\to 2} \, \left(1-\frac{1}{3^{s-1}}\right) \zeta (s)$

which evaluated is:

$\frac{5 \pi ^2}{36} = \frac{\pi ^2}{12} + \frac{\pi ^2}{9}$

but then we would have:

$\frac{5 \pi ^2}{36} = \frac{7 \pi ^2}{36}$

which is not true.

Looking at the logarithm above again, as a zeta function limit, we have:

$\log(n) = \lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$

In this question it was shown in the answer that logarithms have these numerators in their Dirichlet series:

$$1-\text{If}[k \bmod n=0,n,0]$$

which as a matrix is:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

which is an infinite matrix where each row is the numerators in the Dirichlet series that converges to $\log(n)$ for $s=1$.

In the zeta function limit for logarithms the part within the parentheses $\left(1-\frac{1}{n^{s-1}}\right)$ in $\log(n) = \lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$ will go towards zero in the limit:

$\lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) = 0$

In a different matrix:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \end{bmatrix}$$

where $\Lambda(n) = \displaystyle \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$,

and where $\Lambda(n)$ is the von Mangoldt function as proven in the answer to this question, we have the triangle:

$$\displaystyle \begin{bmatrix} +1 \\ +1&-1&=0 \\ +1&+1&-2&=0 \\ +1&-1&+1&-1&=0 \\ +1&+1&+1&+1&-4&=0 \\ +1&-1&-2&-1&+1&+2&=0 \\ +1&+1&+1&+1&+1&+1&-6&=0 \end{bmatrix}$$

where we see that the row sums are also zero.

Looking again at the earlier matrix we see the same thing (with exception of the first row, but $\Lambda(0) = 0$ anyways):

$$\displaystyle \begin{bmatrix} 0=0 \\ 1&-1=0 \\ 1&1&-2=0 \\ 1&1&1&-3=0 \\ 1&1&1&1&-4=0 \\ 1&1&1&1&1&-5=0 \\ 1&1&1&1&1&1&-6=0 \end{bmatrix}$$

writing zeta function limits of this latter matrix, and we have:

$$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}-\frac{1}{2^{s-1}}\right) \zeta (s) = \log (2)$$ $$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}+\frac{1}{2^{s-1}}-\frac{2}{3^{s-1}}\right) \zeta (s) = \log \left(\frac{9}{2}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}+\frac{1}{2^{s-1}}+\frac{1}{3^{s-1}}-\frac{3}{4^{s-1}}\right) \zeta (s) = \log \left(\frac{32}{3}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}+\frac{1}{2^{s-1}}+\frac{1}{3^{s-1}}+\frac{1}{4^{s-1}}-\frac{4}{5^{s-1}}\right) \zeta (s) = \log \left(\frac{625}{24}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}+\frac{1}{2^{s-1}}+\frac{1}{3^{s-1}}+\frac{1}{4^{s-1}}+\frac{1}{5^{s-1}}-\frac{5}{6^{s-1}}\right) \zeta (s) = \log \left(\frac{324}{5}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{1^{s-1}}+\frac{1}{2^{s-1}}+\frac{1}{3^{s-1}}+\frac{1}{4^{s-1}}+\frac{1}{5^{s-1}}+\frac{1}{6^{s-1}}-\frac{6}{7^{s-1}}\right) \zeta (s) = \log \left(\frac{117649}{720}\right)$$

with the sequence: $1, 2, 9/2, 32/3, 625/24, 324/5, 117649/720...$ looking up numerators and denominators in the oeis we find sequences: http://oeis.org/A036505 and: http://oeis.org/A095996

Multiplying by $n!$ and searching for the sequence: $1, 4, 27, 256, 3125, 46656, 823543...$ we find when we scroll down to sequence https://oeis.org/A177885 the following formula:

$$\frac{1}{2}+\frac{2 \pi i \exp (1) \left(n-\frac{11}{8}\right)}{\exp (1) W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)}$$

involving the Lambert_W function for a good approximation of the n-th Riemann zeta zero. $W(x)$ or LambertW is undocumented in Mathematica as said in Mathworld, and the ProductLog[x] command is used instead. There is also a functional relation between the sequence from the zeta limits and the $1, 4, 27, 256, 3125, 46656, 823543...$ sequence which is as follows:

$$\frac{x}{W(-x)} = x \sum _{n=1}^{\infty } \frac{1}{(W(-x)+1)^n}$$

Changing the argument $-x$ to $x$ we have the function used in the zeta zero approximation formula.

This warrants some further investigation of the sequence $1, 2, 9/2, 32/3, 625/24, 324/5, 117649/720...$ from the zeta function limits.

Setting the denominators equal to $\exp(n)^{s-1}$:

$$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}-\frac{1}{\exp ^{s-1}(2)}\right) \zeta (s) =1$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{1}{\exp ^{s-1}(2)}-\frac{2}{\exp ^{s-1}(3)}\right) \zeta (s)=3$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{1}{\exp ^{s-1}(2)}+\frac{1}{\exp ^{s-1}(3)}-\frac{3}{\exp ^{s-1}(4)}\right) \zeta (s)=6$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{1}{\exp ^{s-1}(2)}+\frac{1}{\exp ^{s-1}(3)}+\frac{1}{\exp ^{s-1}(4)}-\frac{4}{\exp ^{s-1}(5)}\right) \zeta (s)=10$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{1}{\exp ^{s-1}(2)}+\frac{1}{\exp ^{s-1}(3)}+\frac{1}{\exp ^{s-1}(4)}+\frac{1}{\exp ^{s-1}(5)}-\frac{5}{\exp ^{s-1}(6)}\right) \zeta (s)=15$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{1}{\exp ^{s-1}(2)}+\frac{1}{\exp ^{s-1}(3)}+\frac{1}{\exp ^{s-1}(4)}+\frac{1}{\exp ^{s-1}(5)}+\frac{1}{\exp ^{s-1}(6)}-\frac{6}{\exp ^{s-1}(7)}\right) \zeta (s)=21$$

which appear to be the triangular numbers, $\frac{n(n+1)}{2}$.

Reversing the triangular matrix: $$\displaystyle \begin{bmatrix} 0=0 \\ 1&-1=0 \\ 1&1&-2=0 \\ 1&1&1&-3=0 \\ 1&1&1&1&-4=0 \\ 1&1&1&1&1&-5=0 \\ 1&1&1&1&1&1&-6=0 \end{bmatrix}$$ into: $$\displaystyle \begin{bmatrix} 0=0 \\ -1&1=0 \\ -2&1&1=0 \\ -3&1&1&1=0 \\ -4&1&1&1&1=0 \\ -5&1&1&1&1&1=0 \\ -6&1&1&1&1&1&1=0 \end{bmatrix}$$ which is the triangle https://oeis.org/A167407 in the OEIS. Taking successive matrix powers of this triangle and reversing back we seem to get the whole Pascal triangle as zeta function limits.

$$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}-\frac{1}{\exp ^{s-1}(2)}\right) \zeta (s)=1$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{2}{\exp ^{s-1}(2)}-\frac{3}{\exp ^{s-1}(3)}\right) \zeta (s)=4$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{2}{\exp ^{s-1}(2)}+\frac{3}{\exp ^{s-1}(3)}-\frac{6}{\exp ^{s-1}(4)}\right) \zeta (s)=10$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{2}{\exp ^{s-1}(2)}+\frac{3}{\exp ^{s-1}(3)}+\frac{4}{\exp ^{s-1}(4)}-\frac{10}{\exp ^{s-1}(5)}\right) \zeta (s)=20$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{2}{\exp ^{s-1}(2)}+\frac{3}{\exp ^{s-1}(3)}+\frac{4}{\exp ^{s-1}(4)}+\frac{5}{\exp ^{s-1}(5)}-\frac{15}{\exp ^{s-1}(6)}\right) \zeta (s)=35$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1)}+\frac{2}{\exp ^{s-1}(2)}+\frac{3}{\exp ^{s-1}(3)}+\frac{4}{\exp ^{s-1}(4)}+\frac{5}{\exp ^{s-1}(5)}+\frac{6}{\exp ^{s-1}(6)}-\frac{21}{\exp ^{s-1}(7)}\right) \zeta (s)=56$$

More in line with logarithms of $n=1,2,3,4,5...$ is to consider the following zeta function limits:

$$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+2)}-\frac{1}{\exp ^{s-1}(2+2)}\right) \zeta (s)=1$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+3)}+\frac{1}{\exp ^{s-1}(2+3)}-\frac{2}{\exp ^{s-1}(3+3)}\right) \zeta (s)=3$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+4)}+\frac{1}{\exp ^{s-1}(2+4)}+\frac{1}{\exp ^{s-1}(3+4)}-\frac{3}{\exp ^{s-1}(4+4)}\right) \zeta (s)=6$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+5)}+\frac{1}{\exp ^{s-1}(2+5)}+\frac{1}{\exp ^{s-1}(3+5)}+\frac{1}{\exp ^{s-1}(4+5)}-\frac{4}{\exp ^{s-1}(5+5)}\right) \zeta (s)=10$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+6)}+\frac{1}{\exp ^{s-1}(2+6)}+\frac{1}{\exp ^{s-1}(3+6)}+\frac{1}{\exp ^{s-1}(4+6)}+\frac{1}{\exp ^{s-1}(5+6)}-\frac{5}{\exp ^{s-1}(6+6)}\right) \zeta (s)=15$$ $$\lim_{s\to 1} \, \left(\frac{1}{\exp ^{s-1}(1+7)}+\frac{1}{\exp ^{s-1}(2+7)}+\frac{1}{\exp ^{s-1}(3+7)}+\frac{1}{\exp ^{s-1}(4+7)}+\frac{1}{\exp ^{s-1}(5+7)}+\frac{1}{\exp ^{s-1}(6+7)}-\frac{6}{\exp ^{s-1}(7+7)}\right) \zeta (s)=21$$

where we get the triangular numbers again. In other words there seems to be an invariance property.

Converting back to unexponentiated denominators:

$$\lim_{s\to 1} \, \left(\frac{1}{(1+2)^{s-1}}-\frac{1}{(2+2)^{s-1}}\right) \zeta (s)=\log \left(\frac{4}{3}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{(1+3)^{s-1}}+\frac{1}{(2+3)^{s-1}}-\frac{2}{(3+3)^{s-1}}\right) \zeta (s)=\log \left(\frac{9}{5}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{(1+4)^{s-1}}+\frac{1}{(2+4)^{s-1}}+\frac{1}{(3+4)^{s-1}}-\frac{3}{(4+4)^{s-1}}\right) \zeta (s)=\log \left(\frac{256}{105}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{(1+5)^{s-1}}+\frac{1}{(2+5)^{s-1}}+\frac{1}{(3+5)^{s-1}}+\frac{1}{(4+5)^{s-1}}-\frac{4}{(5+5)^{s-1}}\right) \zeta (s)=\log \left(\frac{625}{189}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{(1+6)^{s-1}}+\frac{1}{(2+6)^{s-1}}+\frac{1}{(3+6)^{s-1}}+\frac{1}{(4+6)^{s-1}}+\frac{1}{(5+6)^{s-1}}-\frac{5}{(6+6)^{s-1}}\right) \zeta (s)=\log \left(\frac{1728}{385}\right)$$ $$\lim_{s\to 1} \, \left(\frac{1}{(1+7)^{s-1}}+\frac{1}{(2+7)^{s-1}}+\frac{1}{(3+7)^{s-1}}+\frac{1}{(4+7)^{s-1}}+\frac{1}{(5+7)^{s-1}}+\frac{1}{(6+7)^{s-1}}-\frac{6}{(7+7)^{s-1}}\right) \zeta (s)=\log \left(\frac{117649}{19305}\right)$$

where we have: $1,4/3, 9/5, 256/105, 625/189, 1728/385, 117649/19305,...$ compared to the earlier: $1, 2, 9/2, 32/3, 625/24, 324/5, 117649/720...$ in other words, no invariance like with the triangular numbers.

Drawing this further we generalize and write:

$$\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$ $$\lim_{s\to 1} \, \zeta (s) \sum _{k=n+1}^{n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$ $$\lim_{s\to 1} \, \zeta (s) \sum _{k=n+n+1}^{n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$ $$\lim_{s\to 1} \, \zeta (s) \sum _{k=n+n+n+1}^{n+n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$ $$\lim_{s\to 1} \, \zeta (s) \sum _{k=n+n+n+n+1}^{n+n+n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

for which we get the sequences for $n=1,2,3,4,5,6,7,...$:

$$\left\{0,\log (2),\log \left(\frac{9}{2}\right),\log \left(\frac{32}{3}\right),\log \left(\frac{625}{24}\right),\log \left(\frac{324}{5}\right),\log \left(\frac{117649}{720}\right),...\right\}$$ $$\left\{0,\log \left(\frac{4}{3}\right),\log \left(\frac{9}{5}\right),\log \left(\frac{256}{105}\right),\log \left(\frac{625}{189}\right),\log \left(\frac{1728}{385}\right),\log \left(\frac{117649}{19305}\right),...\right\}$$ $$\left\{0,\log \left(\frac{6}{5}\right),\log \left(\frac{81}{56}\right),\log \left(\frac{96}{55}\right),\log \left(\frac{16875}{8008}\right),\log \left(\frac{19683}{7735}\right),\log \left(\frac{3176523}{1033600}\right),...\right\}$$ $$\left\{0,\log \left(\frac{8}{7}\right),\log \left(\frac{72}{55}\right),\log \left(\frac{2048}{1365}\right),\log \left(\frac{5000}{2907}\right),\log \left(\frac{331776}{168245}\right),\log \left(\frac{15059072}{6660225}\right),...\right\}$$ $$\left\{0,\log \left(\frac{10}{9}\right),\log \left(\frac{225}{182}\right),\log \left(\frac{4000}{2907}\right),\log \left(\frac{390625}{255024}\right),\log \left(\frac{4500}{2639}\right),\log \left(\frac{367653125}{193666176}\right),...\right\}$$

What is the relationship between these numbers? Trying to describe the fractions within the logarithms I wrote:

$$\frac{((j+1) n)^j}{\prod _{k=1}^j ((j+1) (n-1)+k)}$$ $$n=1,2,3,4,5$$ $$j=0,1,2,3,4,5,6$$

which gives:

$$\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 2 & \frac{4}{3} & \frac{6}{5} & \frac{8}{7} & \frac{10}{9} \\ \frac{9}{2} & \frac{9}{5} & \frac{81}{56} & \frac{72}{55} & \frac{225}{182} \\ \frac{32}{3} & \frac{256}{105} & \frac{96}{55} & \frac{2048}{1365} & \frac{4000}{2907} \\ \frac{625}{24} & \frac{625}{189} & \frac{16875}{8008} & \frac{5000}{2907} & \frac{390625}{255024} \\ \frac{324}{5} & \frac{1728}{385} & \frac{19683}{7735} & \frac{331776}{168245} & \frac{4500}{2639} \\ \frac{117649}{720} & \frac{117649}{19305} & \frac{3176523}{1033600} & \frac{15059072}{6660225} & \frac{367653125}{193666176} \end{array} \right)$$

which are the sequences within the logarithms above but transposed. Here I tried to permute indexes $j$,$n$ and $k$ in an attempt to transpose the matrix but I did not succeed.

In addition to the known (conjectured):

$$\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}=\frac{n^{n-1}}{(n-1)!}$$

a oeis search and some manipulation gives the following:

$$\lim_{s\to 1} \, \zeta (s) \sum _{k=n+1}^{n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}=\frac{n^{n-1}}{\frac{(n-1)! (2 n-1)\text{!!}}{n!}}$$

The following appears to be an example of a relationship:

$$\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}=\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^{n+n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}+\lim_{s\to 1} \, \zeta (s) \sum _{k=n+1}^{n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}+\lim_{s\to 1} \, \zeta (s) \sum _{k=n+n+1}^{n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}+\lim_{s\to 1} \, \zeta (s) \sum _{k=n+n+n+1}^{n+n+n+n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

for $n$ an integer $\geq 1$. Also the limit $s \to 1$ seems be possible to put to any complex number $S=a+i b$, $s \to S$.

So the question repeated is:

Does: $$\sum _{m=1}^t \lim_{s\to \text{S}} \, \zeta (s) \sum _{k=(m-1) n+1}^{m n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

equal:

$$\lim_{s\to \text{S}} \, \zeta (s) \sum _{k=1}^{n t} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}$$

for any complex number $S=a+i b$ and integer values of $n \geq 1$ and $t \geq 1$?

The question seems to simplify to the following identity $$\sum _{m=1}^t \lim_{s\to \text{S}} \, \zeta (s) \sum _{k=(m-1) n+1}^{m n} \frac{1}{k^{s-1}}=\lim_{s\to \text{S}} \, \zeta (s) \sum _{k=1}^{n t} \frac{1}{k^{s-1}}$$

for $S$ a complex number not equal to $1$ and $n=1,2,3,4,5,...$ and $t=1,2,3,4,5,...$

S = 1.23456789 + I*9.87654321
Monitor[Table[
Chop[N[Table[
Sum[Limit[Zeta[s]*Sum[1/k^(s - 1), {k, (m - 1)*n + 1, m*n}],
s -> S], {m, 1, t}], {n, 1, 12}]] -
Table[Limit[Zeta[s]*Sum[1/k^(s - 1), {k, 1, t*n}], s -> S], {n, 1,
12}]], {t, 1, 6}], t]


Simpler still:

$$\sum _{m=1}^t \left(\sum _{k=(m-1) n+1}^{m n} \frac{1}{k^{s-1}}\right)=\sum _{k=1}^{n t} \frac{1}{k^{s-1}}$$

for $S=a+i b$ and $n=1,2,3,4,5,..$ $t=1,2,3,4,5,..$

Can anyone simplify the indexes in the double sum?

• Looks more like a paper than a question...+1 – draks ... Sep 15 '13 at 18:05

Starting from $$\sum _{m=1}^t \lim_{s\to \text{S}} \, \zeta (s) \sum _{k=(m-1) n+1}^{m n} \frac{1}{k^{s-1}}=\lim_{s\to \text{S}} \, \zeta (s) \sum _{k=1}^{n t} \frac{1}{k^{s-1}}$$ you don't need to use limits, since $\lim_{s \to S} \zeta (s) = \zeta (S)$ and you can divide by $\zeta(S)$ on both sides.
$$\sum _{m=1}^t \sum _{k=(m-1) n+1}^{m n} \frac{1}{k^{S-1}}=\sum _{k=1}^{n t} \frac{1}{k^{S-1}}$$
Now, the inner summation of the left-hand side goes from $1$ to $n$, then from $n+1$ to $2n$, then from $2n+1$ to $3n$ and so on up to $(t-1)n+1$ to $nt$. Since the indices are non-overlapping and consecutive you can conclude that equality holds.