The Zeta Function Limits

C o n c e p t s

This topic requires familiarity with the following concepts:

  • Sequences and Series
  • Riemann Zeta Function
  • Harmonic Series
  • Hyperbolic functions
  • Euler's Series of π
  • Geometric Series

The Riemann zeta function is a special function whose importance is referenced at Mathworld. This topic topic ties in with Euler's Series of π toward the end. I partly claim the discovery below of the zeta limits discussed below. I have only found 1 reference to the values of 3 of the limits, but have not seen a generalization of the limits.

Defining the Zeta Limits

The Zeta function is defined as: $\zeta (s)=1+\frac{1}{{{2}^{s}}}+\frac{1}{{{3}^{s}}}+\frac{1}{{{4}^{s}}}+\frac{1}{{{5}^{s}}}+\frac{1}{{{6}^{s}}}+\frac{1}{{{7}^{s}}}+\frac{1}{{{8}^{s}}}+....=\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{s}}}}$.

The summation of the Zeta values ($\underset{k}{\mathop{\sum }}\,\zeta (k)$) increase approximately at a linear rate if k increases at a constant rate. This can be deduced from the fact that $\underset{s\to \infty }{\mathop{\lim }}\,\zeta (s)=1$. This suggests the possibility that $\underset{n\to \infty }{\mathop{\lim }}\,n-\underset{k}{\overset{n}{\mathop{\sum }}}\,\zeta (k)$ approaches a limit L. This limit can also be written as: $\underset{n\to \infty }{\mathop{\lim }}\,\underset{k}{\overset{n}{\mathop{\sum }}}\,1-\zeta (k)$. I have determined the method for finding limits for certain k. I found the following 3 basic limits at Mathworld but have not seen a generalization of the limits that are present here.

$\underset{n\to \infty }{\mathop{\lim }}\,n-\sum\limits_{k=2}^{n}{\zeta (k)}=-1$     $\underset{n\to \infty }{\mathop{\lim }}\,n-\sum\limits_{k=1}^{n}{\zeta (2k)}=-\frac{3}{4}$    $\underset{n\to \infty }{\mathop{\lim }}\,n-\sum\limits_{k=1}^{n}{\zeta (2k+1)}=-\frac{1}{4}$

So, let's proceed with the proofs of these formulas and the generalization of the limits.

Derivation of Zeta Limits Summation Formula

For a general formula for s increasing at a constant rate, we vary s with u + vi: $\underset{n\to \infty }{\mathop{\lim }}\,n-\sum\limits_{i=1}^{n}{\zeta (u+vi)}$ , where u + v > 1 and v > 0. Since the Harmonic Series does not converge, we have the restriction u + v > 0. Also, it is easy to see that we cannot let v = 0. Hence, we'll think of the limit as a function of u and v. Evaluating the limit for various u and v gives some fascinating results. We’ll define the Zeta function limits or Zeta limits as follows:

(i) $Z(u,v)=\underset{n\to \infty }{\mathop{\lim }}\,n-\sum\limits_{i=1}^{n}{\zeta (u+vi)}$.

This limit can be evaluated by the following summation:

(ii) $Z(u,v)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}({{R}^{v}}-1)}}$.

We will call this the Zeta Limit summation. It is easy to derive this summation. We can think of the limit as a summation of a matrix: $\underset{n\to \infty }{\mathop{\lim }}\,\underset{k}{\overset{n}{\mathop{\sum }}}\,1-\zeta (k)=$ $1-\left( 1+\frac{1}{{{2}^{u+v}}}+\frac{1}{{{3}^{u+v}}}+\frac{1}{{{4}^{u+v}}}+\frac{1}{{{5}^{u+v}}}+\frac{1}{{{6}^{u+v}}}+\frac{1}{{{7}^{u+v}}}+\frac{1}{{{8}^{u+v}}}+... \right)+$
$1-\left( 1+\frac{1}{{{2}^{u+2v}}}+\frac{1}{{{3}^{u+2v}}}+\frac{1}{{{4}^{u+2v}}}+\frac{1}{{{5}^{u+2v}}}+\frac{1}{{{6}^{u+2v}}}+\frac{1}{{{7}^{u+2v}}}+\frac{1}{{{8}^{u+2v}}}+... \right)+$
$1-\left( 1+\frac{1}{{{2}^{u+3v}}}+\frac{1}{{{3}^{u+3v}}}+\frac{1}{{{4}^{u+3v}}}+\frac{1}{{{5}^{u+3v}}}+\frac{1}{{{6}^{u+3v}}}+\frac{1}{{{7}^{u+3v}}}+\frac{1}{{{8}^{u+3v}}}+... \right)+$
$1-\left( 1+\frac{1}{{{2}^{u+4v}}}+\frac{1}{{{3}^{u+4v}}}+\frac{1}{{{4}^{u+4v}}}+\frac{1}{{{5}^{u+4v}}}+\frac{1}{{{6}^{u+4v}}}+\frac{1}{{{7}^{u+4v}}}+\frac{1}{{{8}^{u+4v}}}+... \right)+$
.....

Notice that the 1’s cancel out. We can rearrange this matrix sum as follows:

$-\left( \frac{1}{{{2}^{u+v}}}+\frac{1}{{{2}^{u+2v}}}+\frac{1}{{{2}^{u+3v}}}+\frac{1}{{{2}^{u+4v}}}+\frac{1}{{{2}^{u+5v}}}+\frac{1}{{{2}^{u+6v}}}+\frac{1}{{{2}^{u+7v}}}+\frac{1}{{{2}^{u+8v}}}+... \right)+$
$-\left( \frac{1}{{{3}^{u+v}}}+\frac{1}{{{3}^{u+2v}}}+\frac{1}{{{3}^{u+3v}}}+\frac{1}{{{3}^{u+4v}}}+\frac{1}{{{3}^{u+5v}}}+\frac{1}{{{3}^{u+6v}}}+\frac{1}{{{3}^{u+7v}}}+\frac{1}{{{3}^{u+8v}}}+... \right)+$
$-\left( \frac{1}{{{4}^{u+v}}}+\frac{1}{{{4}^{u+2v}}}+\frac{1}{{{4}^{u+3v}}}+\frac{1}{{{4}^{u+4v}}}+\frac{1}{{{4}^{u+5v}}}+\frac{1}{{{4}^{u+6v}}}+\frac{1}{{{4}^{u+7v}}}+\frac{1}{{{4}^{u+8v}}}+... \right)+$
$-\left( \frac{1}{{{5}^{u+v}}}+\frac{1}{{{5}^{u+2v}}}+\frac{1}{{{5}^{u+3v}}}+\frac{1}{{{5}^{u+4v}}}+\frac{1}{{{5}^{u+5v}}}+\frac{1}{{{5}^{u+6v}}}+\frac{1}{{{5}^{u+7v}}}+\frac{1}{{{5}^{u+8v}}}+... \right)+$ .....

These are all geometric series that can be summed as follows:

$-\frac{1}{{{2}^{u}}({{2}^{v}}-1)}-\frac{1}{{{3}^{u}}({{3}^{v}}-1)}-\frac{1}{{{4}^{u}}({{4}^{v}}-1)}-\frac{1}{{{5}^{u}}({{5}^{v}}-1)}-\frac{1}{{{6}^{u}}({{6}^{v}}-1)}-\frac{1}{{{7}^{u}}({{7}^{v}}-1)}-\frac{1}{{{8}^{u}}({{8}^{v}}-1)}+...$
$=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}({{R}^{v}}-1)}}$

With the above formula, we have converted the limit into a sum. Although there is probably no one formula for the summation of the above, we can evaluate the sum for certain values of u and v. In the sections that follow, I will provide methods of determining the exact summations for certain values of u and v.

Summation of Z(1,1)

The easiest limit to evaluate is Z(1,1).

$Z(1,1)=-\sum\limits_{R=2}^{\infty }{\frac{1}{R(R-1)}}=-\left( \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\frac{1}{6\cdot 7}+... \right)=$
$-\left[ \left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{1}{4}-\frac{1}{5} \right)+\left( \frac{1}{5}-\frac{1}{6} \right)+\left( \frac{1}{6}-\frac{1}{7} \right)+... \right]=-1$

Therefore,

Z(1,1) = $-\sum\limits_{R=2}^{\infty }{\frac{1}{R(R-1)}}$ = –1.

Summations of Type: Z(u,1)

We now focus on the summations where u is general and v = 1. We have already found Z(1,1), which is an exception to the case. We know that $Z(u,1)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}(R-1)}}$. Now, we can decompose $\frac{1}{{{R}^{u}}(R-1)}$ as:

(iii) $\frac{1}{{{R}^{u}}(R-1)}=\frac{1}{{{R}^{u-1}}(R-1)}-\frac{1}{{{R}^{u}}}$.

This is a recursive formula!! From this, we obtain:

(iv) $-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}(R-1)}}=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u-1}}(R-1)}}+\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}}}=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u-1}}(R-1)}}+[\zeta (u)-1]$.

We can write this in a simple form as:

$Z(u,1)=Z(u-1,1)+\zeta (u)-1$ (for u > 1)

If we can find Z(1,1), then we can find Z(u,1) for any u. We already know Z(1,1) = -1. Therefore, we can evaluate Z(2,1) as follows: $Z(2,1)=Z(1,1)+\zeta (2)-1=-2+\zeta (2)$.

Now, we can find Z(3,1): $Z(3,1)=Z(2,1)+\zeta (3)-1=-2+\zeta (2)-1+\zeta (3)=-3+\zeta (2)+\zeta (3)$. So, we have $Z(3,1)=-3+\zeta (2)+\zeta (3)$.

The recursive formula gives the following general formula:

$Z(u,1)=-u+\sum\limits_{k=2}^{u}{\zeta (k)}$ (for u > 1) and Z(1,1) = –1.

Summations of Type: Z(u,2)

We will have to examine two cases for this type of sum: for even u and for odd u. For this type of sum, we have $Z(u,2)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{u}}({{R}^{2}}-1)}}$. Again, just as we decomposed the rational expression for Z(u,1), we can decompose this rational expression as:

(v) $\frac{1}{{{R}^{u}}({{R}^{2}}-1)}=\frac{1}{{{R}^{u-2}}({{R}^{2}}-1)}-\frac{1}{{{R}^{u}}}$.

This gives rise to the recursive formula:

(vi) $Z(u,2)=Z(u-2,2)+\zeta (u)-1$ (for u > 1).

The exception to the case is when u = 1. When u = 1, the sum is: $Z(1,2)=-\sum\limits_{R=2}^{\infty }{\frac{1}{R({{R}^{2}}-1)}}$. We have to be able to evaluate Z(1,2) if we want to be able to use the recursion formula. Although I do not know how to algebraically evaluate this sum, this series sums to –0.25. I am still searching for the proof. Until then, we will assume this as true.

From this, we can find Z(3,2): $Z(3,2)=Z(1,2)+\zeta (3)-1=-\frac{1}{4}+\zeta (3)-1=-\frac{5}{4}+\zeta (3)$. Thus, $Z(3,2)=-\frac{5}{4}+\zeta (3)$. For a general formula, where u is an odd number, we replace u with 2m + 1. Thus,

(vii) $Z(2m+1,2)=-m-\frac{1}{4}+\sum\limits_{k=1}^{m}{\zeta (2k+1)}$ (for m is an integer > 1).

To evaluate the sum for even values of u, we have to be able to sum Z(0,2):

$Z(0,2)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}-1}}=-\sum\limits_{R=2}^{\infty }{\frac{1}{(R-1)(R+1)}}=-\frac{1}{2}\sum\limits_{R=2}^{\infty }{\frac{1}{R-1}-\frac{1}{R+1}}$
$=-\frac{1}{2}\left[ \left( \frac{1}{1}-\frac{1}{3} \right)+\left( \frac{1}{2}-\frac{1}{4} \right)+\left( \frac{1}{3}-\frac{1}{5} \right)+... \right]=-\frac{3}{4}$

Therefore, $Z(0,2)=-\frac{3}{4}$.

To find Z(2,2), we use the recursive formula: $Z(2,2)=Z(0,2)+\zeta (2)-1=-\frac{7}{4}+\zeta (2)$. Thus, $Z(2,2)=-\frac{7}{4}+\zeta (2)$. For a general formula where u is even, we replace u with 2m:

$Z(2m,2)=-m-\frac{3}{4}+\sum\limits_{k=1}^{m}{\zeta (2k)}$ (for m is an integer > 1), and $Z(0,2)=-\frac{3}{4}$.

Summations of Type Z(u,4)

When v = 4, we cannot find the limit sums for all u. However, we can find the sum when u is even. Two different cases occur.

First, let’s find Z(0,4): $Z(0,4)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{4}}-1}}=-\sum\limits_{R=2}^{\infty }{\frac{1}{({{R}^{2}}-1)({{R}^{2}}+1)}}=-\frac{1}{2}\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}-1}-\frac{1}{{{R}^{2}}+1}}$.

We already know that $\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}-1}}=\frac{3}{4}$ and $\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}+1}}=\frac{\pi \cosh \pi -\sinh \pi }{2\cdot \sinh \pi }-\frac{1}{2}$ (see Euler’s Series of Pi for summation of the last expression). From these two formulas, we have:

$Z(0,4)=-\frac{1}{2}\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}-1}-\frac{1}{{{R}^{2}}+1}}=-\frac{5}{8}+\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }$

Now, let’s find Z(2,4): $Z(2,4)=-\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{2}}({{R}^{4}}-1)}}=-\sum\limits_{R=2}^{\infty }{\frac{1}{2}\left( \frac{1}{{{R}^{2}}+1}+\frac{1}{{{R}^{2}}-1} \right)-\frac{1}{{{R}^{2}}}}$
$=-\frac{1}{2}\left( \frac{\pi \cosh \pi -\sinh \pi }{2\cdot \sinh \pi }-\frac{1}{2} \right)-\frac{3}{8}+[\zeta (2)-1]=-\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }-\frac{1}{4}-\frac{3}{8}-1+\zeta (2)$. Therefore,

(xi) $Z(2,4)=-\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }-\frac{9}{8}+\zeta (2)$.

Again, for v = 4, we have the following recursive formula:

(xii) $Z(u,4)=Z(u-4,4)-1+\zeta (u)$

The general formula for Z(4m,4) is given by:

$Z(4m,4)=\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }-m-\frac{5}{8}+\sum\limits_{k=1}^{m}{\zeta (4k)}$ for u is an integer > 0.

The general formula for Z(4m – 2,4) is given by:

$Z(4m-2,4)=-\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }-m-\frac{1}{8}+\sum\limits_{k=1}^{m}{\zeta (4k-2)}$

We cannot find the exact value of Z(1,4), Z(3,4), Z(5,4), etc. where u are odd integers. However, we can find a recursive formula for these as well. We can approximate Z(1,4) and Z(3,4) and use recursion to find the rest. First, $Z(1,4)=-\sum\limits_{R=2}^{\infty }{\frac{1}{R({{R}^{4}}-1)}}=-\frac{1}{2}\sum\limits_{R=2}^{\infty }{\frac{1}{R({{R}^{2}}-1)}}+\frac{1}{2}\sum\limits_{R=2}^{\infty }{\frac{1}{R({{R}^{2}}+1)}}$. Now, let $\vartheta =\sum\limits_{R=2}^{\infty }{\frac{1}{R({{R}^{2}}+1)}}$. Then, $Z(1,4)=-\frac{1}{2}\cdot \frac{1}{4}+\frac{1}{2}\cdot \vartheta =-\frac{1}{8}+\frac{1}{2}\vartheta $, where ϑ is also given by: $\vartheta =\sum\limits_{k=1}^{\infty }{{{(-1)}^{k-1}}\cdot [\zeta (2k+1)-1]}$. The general formula is:

$Z(4m+1,4)=-\frac{1}{8}-m+\frac{1}{2}\vartheta +\sum\limits_{k=1}^{m}{\zeta (4k+1)}$

In a similar manner, if we let $\varphi =\sum\limits_{R=2}^{\infty }{\frac{1}{{{R}^{3}}({{R}^{2}}+1)}}$, then we have the following: $Z(3,4)=-\frac{5}{8}+\frac{\zeta (3)}{2}+\frac{1}{2}\varphi $. The general formula is:

$Z(4m+3,4)=-\frac{5}{8}-m+\frac{1}{2}\varphi +\frac{\zeta (3)}{2}+\sum\limits_{k=1}^{m}{\zeta (4k+3)}$

It is obvious that Z(1,4) + Z(3,4) = Z(3,2). Therefore, $-\frac{1}{8}+\frac{1}{2}\vartheta -\frac{5}{8}+\frac{1}{2}\varphi +\frac{\zeta (3)}{2}=-\frac{5}{4}+\zeta (3)$. This gives us an interesting relationship: $\varphi +\vartheta =\zeta (3)-1$. From this, we can express Z(3,4) as: $Z(3,4)=-\frac{9}{8}+\zeta (3)-\frac{1}{2}\vartheta $. The formula in (xvi) also becomes:

$Z(4m+3,4)=-\frac{9}{8}-m-\frac{1}{2}\vartheta +\zeta (3)+\sum\limits_{k=1}^{m}{\zeta (4k+3)}$

Other Interesting Properties

Some interesting properties are obvious. For example,
  • Z(0,2) + Z(1,2) = Z(1,1)
  • Z(1,2) + Z(2,2) = Z(2,1)
  • Z(2,2) + Z(3,2) = Z(3,1), etc.

Table of Exact Zeta Limit Values


 v
u124
0undefined$-\frac{3}{4}$$-\frac{5}{8}+\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }$
1-1$-\frac{1}{4}$$-\frac{1}{8}+\frac{1}{2}\vartheta $*
2$-2+\zeta (2)$$-\frac{7}{4}+\zeta (2)$$-\frac{9}{8}-\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }+\zeta (2)$
3$-3+\zeta (2)+\zeta (3)$$-\frac{5}{4}+\zeta (3)$$-\frac{9}{8}+\zeta (3)-\frac{1}{2}\vartheta $
4$-4+\zeta (2)+\zeta (3)+\zeta (4)$$-\frac{11}{4}+\zeta (2)+\zeta (4)$$-\frac{13}{8}+\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }+\zeta (4)$
5$-5+\zeta (2)+\zeta (3)+\zeta (4)+\zeta (5)$$-\frac{9}{4}+\zeta (3)+\zeta (5)$$-\frac{9}{8}+\frac{1}{2}\vartheta +\zeta (5)$
6$-6+\zeta (2)+\zeta (3)+\zeta (4)+\zeta (5)+\zeta (6)$$-\frac{15}{4}+\zeta (2)+\zeta (4)+\zeta (6)$$-\frac{17}{8}-\frac{\pi \cosh \pi -\sinh \pi }{4\cdot \sinh \pi }+\zeta (2)+\zeta (6)$
n$-u+\sum\limits_{k=2}^{u}{\zeta (k)}$  
*ϑ ≈ 0.171865985524373...

Table of Approximate Zeta Limit Values


 v
u124
0undefined–0.75–0.086662976266
1–1–0.25–0.03906700723
2–0.355065933147341–0.105065933147341–0.018402956886
3–0.153009029986839–0.0479430968406448–0.008876089602
4–0.0706857962738219–0.0227426994343088–0.004339742555
5–0.0337580411263805–0.0110153416968080–0.00213925209462662
6–0.0164144791368634–0.0053996374448204–0.001059894901621500
7–0.00806570175660946–0.00266606431529796–0.000526812220488176
8–0.00398834558206859–0.00432228124232056–0.000262386356627376
9–0.00197995272887833–0.00057671496280188–0.000130859268542908
10–0.000985377572231053–0.000327706105453897–0.000065319773796995