## Thursday, 19 June 2014

### `Convergence' of Some Divergent Series

‘Convergence’ of Some Divergent Series!

The topic of this article, the idea of attaching a ﬁnite value to divergent series, is no longer a purely mathematical exercise. These ﬁnite values of divergent series have found application in String theory and quantum ﬁeld theory (Casimir eﬀect).

The ﬁnite sum of real/complex numbers is always ﬁnite. The inﬁnite sum of real/complex numbers can be either ﬁnite or inﬁnite. For instance,

$\sum _{k=1}^{n}k=1+2+\dots +n=\frac{n\left(n+1\right)}{2}$

is ﬁnite and

$\sum _{k=1}^{\infty }k=1+2+3+\dots =+\infty .$

If an inﬁnite sum has ﬁnite value it is said to converge, otherwise it is said to diverge. Divergence do not always mean it grows to $±\infty$. For instance,

are diverging, while

is converging. The convergence of a series is deﬁned by the convergence of its partial sum. For instance,

$\sum _{k=1}^{\infty }k$

diverges because its partial sum

${s}_{n}:=\sum _{k=1}^{n}k=\frac{n\left(n+1\right)}{2}$

diverges, as $n\to \infty$. The geometric series

 $\sum _{k=0}^{\infty }{z}^{k}=\frac{1}{1-z},$ (1)

for all $|z|<1$, because its partial sum

${s}_{n}:=\frac{{z}^{n}-1}{z-1}\stackrel{n\to \infty }{\to }\frac{-1}{z-1}.$

Note that the map $T\left(z\right)=\frac{1}{1-z}$ is well-deﬁned between $T:ℂ\setminus \left\{1\right\}\to ℂ$. In fact, $T$ is a holomorphic (complex diﬀerentiable) functions and, hence, analytic. The analytic function $T$ exists for all complex numbers except $z=1$ and, inside the unit disk $\left\{z\in ℂ\mid |z|<1\right\}$, $T$ is same as the power series in (1). Thus, $T$ may be seen as an analytic continuation of

$\sum _{k=0}^{\infty }{z}^{k}$

outside the unit disk, except at $z=1$. A famous unique continuation result from complex analysis says that the analytic continuation is unique. Thus, one may consider the value of $T\left(z\right)$, outside the unit disk, as an ‘extension’ of the divergent series

$\sum _{k=0}^{\infty }{z}^{k}.$

In this sense, setting $z=2$ in $T\left(z\right)$, the divergent series may be seen as taking the ﬁnite value

$\sum _{k=0}^{\infty }{2}^{k}=1+2+4+8+16+\dots =-1.$

Similarly, setting $z=-1$ in $T\left(z\right)$, the oscillating divergent series may be seen as taking the ﬁnite value

$\sum _{k=0}^{\infty }{\left(-1\right)}^{k}=\frac{1}{1-\left(-1\right)}=\frac{1}{2}.$

This sum coincides with the Cesàro sum which is a special kind of convergence for series which do not diverge to $±\infty$. For instance,

$\sum _{k=1}^{\infty }k$

diverges in Cesàro sum too. The Cesàro sum of a series

$\sum _{k=1}^{\infty }{a}_{k}$

is deﬁned as the

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{i=1}^{n}\left(\sum _{k=1}^{i}{a}_{k}\right).$

Observe that the sequence

$\frac{1}{n}\sum _{i=1}^{n}\left(\sum _{k=0}^{i-1}{\left(-1\right)}^{k}\right)\to \frac{1}{2}.$

Since $T$ is not deﬁned on $1$, we have not assigned any ﬁnite value to the divergent series

$1+1+1+\dots .$

The Riemann zeta function

$\zeta \left(z\right):=\sum _{k=1}^{\infty }\frac{1}{{k}^{z}}$

converges for $z\in ℂ$ with $\Re \left(z\right)>1$. Note that for $z\in ℂ$, ${k}^{z}={e}^{zln\left(k\right)}$, where $ln$ is the real logarithm. If $z=\sigma +it$ then ${k}^{z}={k}^{\sigma }{e}^{itln\left(k\right)}$ and $|{k}^{z}|={k}^{\sigma }$ since $|{e}^{itln\left(n\right)}|=1$. Therefore, the inﬁnite series converges for all $z\in ℂ$ such that $\Re \left(z\right)>1$ (i.e. $\sigma >1$) because it is known that ${\sum }_{k=1}^{\infty }{k}^{-\sigma }$ converges for all $\sigma >1$ and diverges for all $\sigma \le 1$.

The Riemann zeta function is a special case of the Dirichlet series

$D\left(z\right):=\sum _{k=1}^{\infty }\frac{{a}_{k}}{{k}^{z}}$

with ${a}_{k}=1$ for all $k$. Note that for $z=2$, we get the Basel series

$\zeta \left(2\right):=\sum _{k=1}^{\infty }\frac{1}{{k}^{2}}=\frac{{\pi }^{2}}{6}$

and $\zeta \left(3\right)=1.202056903159594\dots$ called the Apéry’s constant. In fact, the following result of Euler gives the value of Riemann zeta function for all even positive integers.

Theorem 1 (Euler). For all $k\in ℕ$,

$\zeta \left(2k\right)={\left(-1\right)}^{k+1}\frac{{\left(2\pi \right)}^{2k}}{2\left(2k\right)!}{B}_{2k}$

where ${B}_{2k}$ is the $2k$-th Bernoulli number.

If $\zeta \left(z\right)$ is well-deﬁned for $\Re \left(z\right)>1$ then ${2}^{1-z}\zeta \left(z\right)$ is also well-deﬁned for $\Re \left(z\right)>1$. Therefore,

$\zeta \left(z\right)\left(1-{2}^{1-z}\right)=\sum _{k=1}^{\infty }{\left(-1\right)}^{k-1}{n}^{-z}=:\eta \left(z\right).$

Observe that $\eta \left(z\right)$ is also a Dirichlet series. It can be shown that $\left(1-{2}^{1-z}\right)\eta \left(z\right)$ is analytic for all $\Re \left(z\right)>0$ and $\Re \left(z\right)\ne 1$. Thus, we have analytically continued $\zeta \left(z\right)$ for all $\Re \left(z\right)>0$ and $\Re \left(z\right)\ne 1$. There is a little work to be done on the zeroes of $1-{2}^{1-z}$ but is ﬁxable. In the strip $0<\Re \left(z\right)<1$, called the critical strip, the Riemann zeta function satisﬁes the relation

$\zeta \left(z\right)={2}^{z}{\left(\pi \right)}^{z-1}sin\left(\frac{z\pi }{2}\right)\Gamma \left(1-z\right)\zeta \left(1-z\right).$

This relation is used to extend the Riemann zeta function to $z$ with non-positive real part, thus, extending to all complex number $z\ne 1$. Setting $z=-1$ in the relation yields

$\zeta \left(-1\right)=\frac{1}{2{\pi }^{2}}sin\left(\frac{-\pi }{2}\right)\Gamma \left(2\right)\zeta \left(2\right)=\frac{1}{2{\pi }^{2}}\left(-1\right)\frac{{\pi }^{2}}{6}=\frac{-1}{12}.$

Since $\zeta \left(-1\right)$ is extension of the series

$\sum _{k=1}^{\infty }\frac{1}{{k}^{z}}$

one may think of $\zeta \left(-1\right)$ as the ﬁnite value corresponding to $1+2+3+\dots$. The value of $\zeta \left(0\right)$ is obtained limiting process because it involves $\zeta \left(1\right)$ which is not deﬁned. However, there is an equivalent formulation of Riemann zeta function for all $z\in ℂ\setminus \left\{1\right\}$ as

$\zeta \left(z\right)=\frac{1}{1-{2}^{1-z}}\sum _{k=0}^{\infty }\frac{1}{{2}^{k+1}}\sum _{i=0}^{k}{\left(-1\right)}^{i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(i+1\right)}^{-z}$

with a simple pole and residue $1$ at $z=1$.

Recall that the analytic continuation $T$ of the geometric series is not deﬁned for $z=1$ and, hence, we could not assign a ﬁnite value to

$T\left(1\right)=1+1+1+\dots .$

But, setting $z=0$ above,

$T\left(1\right)=\zeta \left(0\right)=-\sum _{k=0}^{\infty }\frac{1}{{2}^{k+1}}\sum _{i=0}^{k}{\left(-1\right)}^{i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right)=-\sum _{k=0}^{\infty }\frac{{\delta }_{0k}}{{2}^{k+1}}=-\frac{1}{2}$

where ${\delta }_{0k}$ is the Kronecker delta deﬁned as

${\delta }_{0k}=\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & k=0\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & k\ne 0.\hfill \end{array}\right\$

The harmonic series

$\sum _{n=1}^{\infty }\frac{1}{n}$

corresponds to $\zeta \left(1\right)$ which is not deﬁned. The closest one can conclude about $\zeta \left(1\right)$ is that

$\underset{n\to \infty }{lim}\left(\sum _{k=1}^{n}\frac{1}{k}-ln\left(n\right)\right)=\gamma$

where $\gamma$ is the Euler-Mascheroni constant which has the value $\gamma =0.57721566\dots$.