## 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 $\gamma =0.57721566\dots$.

## Tuesday, 17 June 2014

### Bernoulli Numbers and Polynomials

Bernoulli Numbers and Polynomials

The sum of ﬁrst $n$ natural numbers $1,2,3,\dots ,n$ is

${S}_{1}\left(n\right):=\sum _{m=1}^{n}m=\frac{n\left(n+1\right)}{2}=\frac{{n}^{2}}{2}+\frac{n}{2}.$

This formula can be derived by noting that

$\begin{array}{rcll}{S}_{1}\left(n\right)& =& 1+2+\dots +n& \text{}\\ {S}_{1}\left(n\right)& =& n+\left(n-1\right)+\dots +1.& \text{}\end{array}$

Therefore, summing term-by-term,

$2{S}_{1}\left(n\right)=\underset{n-\text{times}}{\underbrace{\left(n+1\right)+\dots +\left(n+1\right)}}=n\left(n+1\right).$

An alternate way of obtaining the above sum is by using the following two identity:

(i)
${\left(m+1\right)}^{2}-{m}^{2}=2m+1$ and, hence,
$2\sum _{m=1}^{n}m=\sum _{m=1}^{n}\left[{\left(m+1\right)}^{2}-{m}^{2}\right]-n.$

(ii)
$\begin{array}{rcll}\sum _{m=1}^{n}\left[{\left(m+1\right)}^{2}-{m}^{2}\right]& =& \left[{2}^{2}-{1}^{2}\right]+\left[{3}^{2}-{2}^{2}\right]+\dots +& \text{}\\ & & +\left[{n}^{2}-{\left(n-1\right)}^{2}\right]+\left[{\left(n+1\right)}^{2}-{n}^{2}\right]& \text{}\\ & =& {\left(n+1\right)}^{2}-1.& \text{}\end{array}$

Thus,

${S}_{1}\left(n\right)=\frac{{\left(n+1\right)}^{2}-\left(n+1\right)}{2}=\frac{n\left(n+1\right)}{2}.$

More generally, the sum of $k$-th power of ﬁrst $n$ natural numbers is denoted as

${S}_{k}\left(n\right):={1}^{k}+{2}^{k}+\dots +{n}^{k}.$

Since ${a}^{0}=1$, for any $a$, we have ${S}_{0}\left(n\right)=n$. For $k\in ℕ$, one can compute ${S}_{k}\left(n\right)$ using the identities:

(i)
${\left(m+1\right)}^{k+1}-{m}^{k+1}=\sum _{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){m}^{i}$

and, hence,

$\left(k+1\right)\sum _{m=1}^{n}{m}^{k}=\sum _{m=1}^{n}\left[{\left(m+1\right)}^{k+1}-{m}^{k+1}\right]-\sum _{i=0}^{k-1}\sum _{m=1}^{n}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){m}^{i}.$

(ii)
${\sum }_{m=1}^{n}\left[{\left(m+1\right)}^{k+1}-{m}^{k+1}\right]={\left(n+1\right)}^{k+1}-1$.

Thus,

$\begin{array}{rcll}\left(k+1\right)\sum _{m=1}^{n}{m}^{k}& =& {\left(n+1\right)}^{k+1}-1-\sum _{i=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right)\sum _{m=1}^{n}{m}^{i}& \text{}\\ {S}_{k}\left(n\right)& =& \frac{{\left(n+1\right)}^{k+1}}{k+1}-\frac{\left(n+1\right)}{k+1}-\frac{1}{k+1}\sum _{i=1}^{k-1}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){S}_{i}\left(n\right).& \text{}\end{array}$

The formula obtained in RHS is a $\left(k+1\right)$-degree polynomial of $n$. Using the above formula, one can compute

$\begin{array}{rcll}{S}_{2}\left(n\right)& =& \frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}+\frac{n}{6},& \text{}\\ {S}_{3}\left(n\right)& =& \frac{{n}^{4}}{4}+\frac{{n}^{3}}{2}+\frac{{n}^{2}}{4},& \text{}\\ {S}_{4}\left(n\right)& =& \frac{{n}^{5}}{5}+\frac{{n}^{4}}{2}+\frac{{n}^{3}}{3}-\frac{n}{30},& \text{}\\ {S}_{5}\left(n\right)& =& \frac{{n}^{6}}{6}+\frac{{n}^{5}}{2}+\frac{5{n}^{4}}{12}-\frac{{n}^{2}}{12},& \text{}\\ {S}_{6}\left(n\right)& =& \frac{{n}^{7}}{7}+\frac{{n}^{6}}{2}+\frac{{n}^{5}}{2}-\frac{{n}^{3}}{6}+\frac{n}{42},& \text{}\\ {S}_{7}\left(n\right)& =& \frac{{n}^{8}}{8}+\frac{{n}^{7}}{2}+\frac{7{n}^{6}}{12}-\frac{7{n}^{4}}{24}+\frac{{n}^{2}}{12},& \text{}\\ \dots & & & \text{}\end{array}$

James/Jacques/Jakob Bernoulli observed that the sum of ﬁrst $n$ whole numbers raised to the $k$-th power can be concisely written as,

${S}_{k}\left(n\right)=\frac{{n}^{k+1}}{k+1}+\frac{1}{2}{n}^{k}+\frac{1}{12}k{n}^{k-1}+0×{n}^{k-2}+\dots .$

Note that the coeﬃcients $1,1∕2,1∕12,0,\dots$ are independent of $k$. Jakob Bernoulli rewrote the above expression as

$\begin{array}{rcll}\left(k+1\right){S}_{k}\left(n\right)& =& {n}^{k+1}-\frac{1}{2}\left(k+1\right){n}^{k}+\frac{1}{6}\frac{k\left(k+1\right)}{2}{n}^{k-1}+0-& \text{}\\ & & -\frac{1}{30}\frac{\left(k-2\right)\left(k-1\right)k\left(k+1\right)}{4!}+\dots & \text{}\end{array}$

and, hence,

 ${S}_{k}\left(n\right)=\frac{1}{k+1}\sum _{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){B}_{i}{n}^{k+1-i},$ (1)

where ${B}_{i}$ are the $i$-th Bernoulli numbers and the formula is called Bernoulli formula. An easier way to grasp the above formula for ${S}_{k}\left(n\right)$ is to rewrite1 it as

${S}_{k}\left(n\right)=\frac{{\left(n+B\right)}^{k+1}-{B}_{k+1}}{k+1}$

where $B$ is a notation used to identify the $i$-th power of $B$ with the $i$-th Bernoulli number ${B}_{i}$ and

${\left(n+B\right)}^{k+1}:=\sum _{i=0}^{k+1}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){B}_{i}{n}^{k+1-i}.$

This notation also motivates the deﬁnition of Bernoulli polynomial of degree $k$ as

${B}_{k}\left(t\right):=\sum _{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){B}_{i}{t}^{k-i}$

where ${B}_{i}$ are the Bernoulli numbers. In terms of Bernoulli polynomials, the $k$-th Bernoulli number ${B}_{k}={B}_{k}\left(0\right)$.

Two quick observation can be made from (1).

(i)
There is no constant term in ${S}_{k}\left(n\right)$ because $i$ does not take $k+1$.
(ii)
The $k$-th Bernoulli number, ${B}_{k}$, is the coeﬃcient of $n$ in ${S}_{k}\left(n\right)$. For instance, ${B}_{0}$ is coeﬃcient of $n$ in ${S}_{0}\left(n\right)=n$ and, hence, ${B}_{0}=1$. Similarly, ${B}_{1}=1∕2,{B}_{2}=1∕6,{B}_{3}=0,{B}_{4}=-1∕30,{B}_{5}=0,{B}_{6}=1∕42,{B}_{7}=0,\dots$.

The beauty about the sequence of Bernoulli numbers is that one can compute them a priori and use it to calculate ${S}_{k}\left(n\right)$, i.e., given $n\in ℕ$ and $k\in ℕ\cup \left\{0\right\}$ it is enough to know ${B}_{i}$, for all $0\le i\le k$, to compute ${S}_{k}\left(n\right)$. We already computed ${B}_{0}=1$. Using $n=1$ in the Bernoulli formula (1), we get

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

and, this implies that the $k$-th Bernoulli number, for $k>0$, is deﬁned as

${B}_{k}=1-\frac{1}{k+1}\sum _{i=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{k+1}{i}\right){B}_{i}.$

Recall that ${B}_{3},{B}_{5},{B}_{7}$ vanish. In fact, it turns out that ${B}_{k}=0$ for all odd $k>1$. The odd indexed Bernoulli numbers vanish because they have no $n$-term. Since there are no constant terms in ${S}_{k}\left(n\right)$, the vanishing of Bernoulli numbers is equivalent to the fact that ${n}^{2}$ is a factor of ${S}_{k}\left(n\right)$. Some properties of Bernoulli numbers:

(i)
For odd $k>1$, ${B}_{k}=0$.
(ii)
For even $k$, ${B}_{k}\ne 0$.
(iii)
${B}_{k}\in ℚ$ for all $k\in ℕ\cup \left\{0\right\}$.
(iv)
${B}_{0}=1$ is the only non-zero integer.
(v)
${B}_{4j}$ is negative rational and ${B}_{4j-2}$ is positive rational, for all $j\in ℕ$.
(vi)
$|{B}_{6}=1∕42|<|{B}_{2k}|$, for all $k\in ℕ$.

L. Euler gave a nice generating function for the Bernoulli numbers. He seeked a function $f\left(x\right)$ such that ${f}^{\left(k\right)}\left(0\right)={B}_{k}$ where ${f}^{\left(k\right)}$ denotes the $k$-th derivative of $f$ with the convention that ${f}^{\left(0\right)}=f$. If such an $f$ exists then it admits the Taylor series expansion, around $0$,

$f\left(x\right)=\sum _{k=0}^{\infty }{f}^{\left(k\right)}\left(0\right)\frac{{x}^{k}}{k!}.$

Therefore, for such a function

$f\left(x\right)=\sum _{k=0}^{\infty }{B}_{k}\frac{{x}^{k}}{k!}.$

Recall the Taylor series of ${e}^{x}$,

${e}^{x}=\sum _{k=0}^{\infty }\frac{{x}^{k}}{k!}$

deﬁned for all $x\in ℝ$. Consider the product (discrete convolution/Cauchy product) of the inﬁnite series

$\begin{array}{rcll}f\left(x\right){e}^{x}& =& \left(\sum _{k=0}^{\infty }{B}_{k}\frac{{x}^{k}}{k!}\right)\left(\sum _{k=0}^{\infty }\frac{{x}^{k}}{k!}\right)& \text{}\\ & =& \sum _{k=0}^{\infty }\left(\sum _{i=0}^{k}{B}_{i}\frac{{x}^{i}}{i!}\frac{{x}^{k-i}}{\left(k-i\right)!}\right)& \text{}\\ & =& \sum _{k=0}^{\infty }\left(\sum _{i=0}^{k}\frac{{B}_{i}}{i!\left(k-i\right)!}\right){x}^{k}& \text{}\\ & =& \sum _{k=0}^{\infty }\left(\sum _{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){B}_{i}\right)\frac{{x}^{k}}{k!}.& \text{}\end{array}$

Let

${c}_{k}:=\sum _{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){B}_{i}=\sum _{i=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){B}_{i}+{B}_{k}=k+{B}_{k}.$

Then

$f\left(x\right){e}^{x}=\sum _{k=0}^{\infty }\left(k+{B}_{k}\right)\frac{{x}^{k}}{k!}=\sum _{k=1}^{\infty }\frac{{x}^{k}}{\left(k-1\right)!}+f\left(x\right)=x{e}^{x}+f\left(x\right).$

Thus, the $f$ we seek satisﬁes

$f\left(x\right)=\frac{x{e}^{x}}{{e}^{x}-1}$

and is called the generating function. Since ${e}^{x}>0$ for all $x\in ℝ$, we may rewrite $f\left(x\right)$ as

 $f\left(x\right)=\frac{x}{1-{e}^{-x}}.$ (3)

The entire exercise of seeking $f$ can be generalised to complex numbers and

A word of caution that idenitities (2) and (3) are diﬀerent from the standard formulae because we have derived them for second Bernoulli numbers, viz., with ${B}_{1}=1∕2$. The standard convention is to work with ﬁrst Bernoulli numbers, viz., with ${B}_{1}=-1∕2$. The ﬁrst Bernoulli numbers can be obtained by following the approach of summing the $k$-th powers of ﬁrst $n-1$ natural numbers, for any given $n$.

The Bernoulli numbers with appeared while computing ${S}_{k}\left(n\right)$ is appears in many crucial places.

(a)
In the expansion of $tanz$.
$tanz=\sum _{k=1}^{\infty }{\left(-1\right)}^{k-1}\frac{{2}^{2k}\left({2}^{2k}-1\right){B}_{2k}}{2n!}{z}^{2k-1}.$

for all $|z|<\pi ∕2$.

(b)
In computing the sum of Riemann-zeta function
$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{{n}^{s}}$

for positive even integers $s$. The case $s=2$ is the famous Basel problem computed by Euler to be ${\pi }^{2}∕6$.

Theorem 1 (Euler). For all $k\in ℕ$,

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

Further, the relation $\zeta \left(-2k\right)=-\frac{{B}_{2k+1}}{2k+1}$ gives the trivial zeroes of the Riemann zeta function.

(c)
The Bernoulli numbers were also used as an attempt to prove Fermat’s last theorem (already discussed in a previous article/blog).

Deﬁnition 1. An odd prime number $p$ is called regular if $p$ does not divide the numerator of ${B}_{k}$, for all even $k\le p-3$. Any odd prime which is not regular is called irregular.

The odd primes $3,5,7,\dots ,31$ are all regular primes. The ﬁrst irregular prime is $37$. It is an open question: are there inﬁnitely many regular primes? However, it is known that there are inﬁnitely many irregular primes.

Theorem 2 (Kummer, 1850). If $p$ is a regular odd prime then the equation

${a}^{p}+{b}^{p}={c}^{p}$

has no solution in $ℕ$.