Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.6.6 The Alternating Harmonic Series and its Inverse Analog

We return to the field SH=DX−1X{S_{H} = D}_{X}^{- 1}X, from the last sec­tion, which was equiv­a­lent to the outer prod­uct of el­e­ments in the two series

SH=(∑k=1∞(−1)k+1k,)⊗(∑k=1∞(−1)k+1k),\begin{array}{r} S_{H} = \left( \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k,} \right) \otimes \left( \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}}{k} \right), \end{array}
(1.6.180)

whose matrix SHS_{H}, we again show, as

SH=DX−1X=(1−1213−1415−1617⋯−21−2312−2513−27⋯3−321−3435−1237⋯−42−431−4523−47⋯5−5253−541−5657⋯−63−232−651−67⋯7−7273−7475−761⋯⋮⋮⋮⋮⋮⋮⋮⋱).\begin{array}{r} {S_{H} = D}_{X}^{- 1}X = \begin{pmatrix} 1 & - \frac{1}{2} & \frac{1}{3} & - \frac{1}{4} & \frac{1}{5} & - \frac{1}{6} & \frac{1}{7} & \cdots \\ - 2 & 1 & - \frac{2}{3} & \frac{1}{2} & - \frac{2}{5} & \frac{1}{3} & - \frac{2}{7} & \cdots \\ 3 & - \frac{3}{2} & 1 & - \frac{3}{4} & \frac{3}{5} & - \frac{1}{2} & \frac{3}{7} & \cdots \\ - 4 & 2 & - \frac{4}{3} & 1 & - \frac{4}{5} & \frac{2}{3} & - \frac{4}{7} & \cdots \\ 5 & - \frac{5}{2} & \frac{5}{3} & - \frac{5}{4} & 1 & - \frac{5}{6} & \frac{5}{7} & \cdots \\ - 6 & 3 & - 2 & \frac{3}{2} & - \frac{6}{5} & 1 & - \frac{6}{7} & \cdots \\ 7 & - \frac{7}{2} & \frac{7}{3} & - \frac{7}{4} & \frac{7}{5} & - \frac{7}{6} & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}. \end{array}
(1.6.181)

As seen in the array above, the in­verse al­ter­nat­ing har­mon­ic series is rep­re­sent­ed in the first row, and the al­ter­nat­ing nat­ur­al num­bers are in the first column, which relate by the equiv­a­lence ex­pres­sion,

∑k=1∞(−1)k+1(1k)−1=∑k=1∞(−1)k+1k.\begin{array}{r} \sum_{k = 1}^{\infty}{( - 1)^{k + 1}\left( \frac{1}{k} \right)^{- 1} = \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k.}} \end{array}
(1.6.182)

We know that the har­mon­ic series di­verges, albeit very slowly, but the al­ter­nat­ing har­mon­ic series con­verges to ln⁡(2)\ln(2)

∑k=1∞(−1)k+1k=1−12+13−14+15−16+…=ln⁡(2),\begin{array}{r} \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots = \ln(2), \end{array}
(1.6.183)

which has par­tial sums equal to

∑k=1n(−1)k+1k=(−1)n+1Φ(−1,1,n+1)+ln⁡(2),\begin{array}{r} \sum_{k = 1}^{n}\frac{( - 1)^{k + 1}}{k} = ( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2), \end{array}
(1.6.184)

where the func­tion Φ(z,s,a)\Phi(z,s,a), is the Lerch tran­scen­dent, de­fined as the sum­ma­tion

Φ(z,s,α)=∑k=0∞zk(α+k)s,\begin{array}{r} \Phi(z,s,\alpha) = \sum_{k = 0}^{\infty}{\frac{z^{k}}{(\alpha + k)^{s}},} \end{array}
(1.6.185)

with the in­te­gral rep­re­sen­ta­tion

Φ(z,s,α)=1Γ(s)∫0∞ts−1e−αt1−ze−t.\begin{array}{r} \Phi(z,s,\alpha) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}{\frac{t^{s - 1}e^{- \alpha t}}{1 - ze^{- t}}.} \end{array}
(1.6.186)

The Φ(z,s,α)\Phi(z,s,\alpha) func­tion en­com­pass­es many famous func­tions that play a cen­tral role in math­e­mat­i­cal analy­sis, and ar­guably the most famous of them is the Rie­mann zeta func­tion ζ(s),\zeta(s),which is equiv­a­lent to Φ(1,s,0)\Phi(1,s,0)

ζ(s)=Φ(1,s,0)=∑k=1∞1ks.\begin{array}{r} \zeta(s) = \Phi(1,s,0) = \sum_{k = 1}^{\infty}{\frac{1}{k^{s}}.} \end{array}
(1.6.187)

Plot­ting par­tial sums, SkS_{k}, for the func­tion (−1)n+1Φ(−1,1,n+1)+ln⁡(2)( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2) in Figure 1.6.15, we can see that these are trend­ing to the con­ver­gence value of ln⁡(2)≈0.6931\ln(2) \approx 0.6931.

Figure
Figure 1.6.15 – Par­tial sums SkS_{k}, of (−1)n+1Φ(−1,1,n+1)+ln⁡(2)( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2).

The par­tial sum of the al­ter­nat­ing har­mon­ic series, can also be ex­pressed in the form of the Lerch tran­scen­dent and ln⁡(2)\ln(2)

∑k=1n(−1)k+1k=(−1)n+1Φ(−1,1,n+1)+ln⁡(2),\begin{array}{r} \sum_{k = 1}^{n}\frac{( - 1)^{k + 1}}{k} = ( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2), \end{array}
(1.6.188)

and it is in­trigu­ing that the sum of two in­fi­nite series con­verges to our finite par­tial AHS sum, where the Lerch tran­scen­dent is de­fined, as Φ(−1,1,n+1)\Phi( - 1,1,n + 1)

Φ(−1,1,n+1)=∑k=0∞(−1)k(n+1+k)1,\begin{array}{r} \Phi( - 1,1,n + 1) = \sum_{k = 0}^{\infty}{\frac{( - 1)^{k}}{(n + 1 + k)^{1}},} \end{array}
(1.6.189)

and our par­tial sum func­tion equals

(−1)n+1Φ(−1,1,n+1)+ln⁡(2)=(∑k=0∞(−1)n+1+k(n+1+k)1)+ln⁡(2).\begin{array}{r} ( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2) = \left( \sum_{k = 0}^{\infty}\frac{( - 1)^{n + 1 + k}}{(n + 1 + k)^{1}} \right) + \ln(2). \end{array}
(1.6.190)

Ex­pand­ing this sum­ma­tion

(∑k=0∞(−1)n+1+k(n+1+k)1)+ln⁡(2)=(−1)n+1(n+1)1+(−1)n+2(n+2)1+(−1)n+3(n+3)1+…+(−1)n+∞(n+∞)1+ln⁡(2),\begin{array}{r} \left( \sum_{k = 0}^{\infty}\frac{( - 1)^{n + 1 + k}}{(n + 1 + k)^{1}} \right) + \ln(2) = \frac{( - 1)^{n + 1}}{(n + 1)^{1}} + \frac{( - 1)^{n + 2}}{(n + 2)^{1}} + \frac{( - 1)^{n + 3}}{(n + 3)^{1}} + \ldots + \frac{( - 1)^{n + \infty}}{(n + \infty)^{1}} + \ln(2), \end{array}
(1.6.191)

it is easy to see the log­a­rithm func­tion in the sum­ma­tion terms. The Lerch tran­scen­dent ex­tends the al­ter­nat­ing har­mon­ic series to the com­plex plane and beyond the "zero-point", by taking the par­tial sums of the al­ter­nat­ing har­mon­ic series and making the func­tion con­tin­u­ous, as seen in the figure below.

Figure
Figure 1.6.16 –Plot of (−1)n+1Φ(−1,1,n+1)+ln⁡(2)( - 1)^{n + 1}\Phi( - 1,1,n + 1) + \ln(2).

The now fa­mil­iar in­verse har­mon­ic series or al­ter­nat­ing nat­ur­al num­bers,

∑k=1∞(−1)k+1k,\begin{array}{r} \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k,} \end{array}
(1.6.192)

has a closed form for its par­tial sums Sj,S_{j},in the form

∑j=1k(−1)j+1j=14((−1)k+1−2k(−1)k+1),\begin{array}{r} \sum_{j = 1}^{k}{( - 1)^{j + 1}j = \frac{1}{4}\left( ( - 1)^{k + 1} - 2k( - 1)^{k} + 1 \right),} \end{array}
(1.6.193)

but for the limit, as n→∞n \rightarrow \infty, SjS_{j} ap­pears to be di­ver­gent

limn→∞⁡(14((−1)n+1−2n(−1)n+1))=∞.\begin{array}{r} \lim_{n \rightarrow \infty}\left( \frac{1}{4}\left( ( - 1)^{n + 1} - 2n( - 1)^{n} + 1 \right) \right) = \infty. \end{array}
(1.6.194)

It turns out the series is reg­u­lar­iz­able, where­by using the Borel reg­u­lar­iza­tion tech­nique, we find 14\frac{1}{4} for its result

lims→1⁡∫0∞e−st(∑k=0∞(−1)k+2(k+1)tkk!) dt=lims→1⁡∫0∞e−st(−e−t(t−1)) dt\lim_{s \rightarrow 1}{\int_{0}^{\infty}{e^{- st}\left( \sum_{k = 0}^{\infty}\frac{( - 1)^{k + 2}(k + 1)t^{k}}{k!} \right)}}\ dt = \lim_{s \rightarrow 1}{\int_{0}^{\infty}{e^{- st}\left( - e^{- t}(t - 1) \right)}}\ dt
=∫0∞−e−2t(t−1) dt= e−2t(2t−1)4|0∞=14.\begin{array}{r} = \int_{0}^{\infty}{- e^{- 2t}(t - 1)}\ dt = \left. \ \frac{e^{- 2t}(2t - 1)}{4} \right|_{0}^{\infty} = \frac{1}{4}. \end{array}
(1.6.195)

Like the har­mon­ic num­bers HkH_{k}, the in­verse har­mon­ic num­bers are com­posed of par­tial sums of the nat­ur­al number set ℕ\mathbb{N}. We choose the no­ta­tion NkN_{k}, to rep­re­sent par­tial sums of the nat­ur­al num­bers, and Hk±H_{k}^{\pm} for par­tial sums of the al­ter­nat­ing har­mon­ic series. And a simple closed form also exists for NkN_{k}, as

Nk=k(k+1)2.\begin{array}{r} N_{k} = \frac{k(k + 1)}{2}. \end{array}
(1.6.196)

Fur­ther defin­ing par­tial sums of al­ter­nat­ing nat­ur­al num­bers as Nk±N_{k}^{\pm}, they have the very beau­ti­ful prop­er­ty of os­cil­lat­ing be­tween pos­i­tive and neg­a­tive values for each suc­ces­sive it­er­a­tion in their par­tial sum series. By ac­cept­ing that kk is not re­strict­ed to the nat­ur­al num­bers, and in­stead that k∈ℝk\in \mathbb{R}, we find the ap­peal­ing plot in Figure 1.6.17. The image shows the real valued por­tion of the plot forms a sym­met­ric wave­form em­a­nat­ing out from the value k=−12k = - \frac{1}{2}, while the imag­i­nary values also os­cil­late, but the wave­form is not sym­met­ric, acting like an odd func­tion. Both of the wave­forms will main­tain con­sis­tent wave­lengths, while steadi­ly grow­ing or shrink­ing in am­pli­tude, de­pend­ing on their move­ment toward or from the value k=0.k = 0.Given its wave­like ap­pear­ance, it is not sur­pris­ing that Nk±N_{k}^{\pm}, for k∈ℝk\in \mathbb{R}, can be ex­pressed using sine and cosine func­tions, as

f(Nk )=(1+2k)4(cos⁡(kπ)+isin(kπ))−14.\begin{array}{r} f\left( N_{k}\ \right) = \frac{(1 + 2k)}{4}\left( \cos(k\pi) + isin(k\pi) \right) - \frac{1}{4}. \end{array}
(1.6.197)
Figure
Figure 1.6.17 –Plot - al­ter­nat­ing in­verse har­mon­ic num­bers, Nk±=14((1+2k)(cos⁡(kπ)+isin(kπ))+1)N_{k}^{\pm} = \frac{1}{4}\left( (1 + 2k)\left( \cos(k\pi) + isin(k\pi) \right) + 1 \right), k∈ℝ.k\in \mathbb{R}.

Zoom­ing in on the figure above to the area just to the left of the origin, re­veals a very beau­ti­ful part of the func­tion (Figure 1.6.18), where at­ten­tion should be di­rect­ed to the point k=−12,k = - \frac{1}{2},which solv­ing the par­tial sum equa­tion for the in­verse har­mon­ic number N(−12)±N_{\left( - \frac{1}{2} \right)}^{\pm}, we find that it equals our Borel reg­u­lar­iza­tion value of 14,\frac{1}{4},

 N(−12)±=14((−1)12−2(−12)(−1)12+1)=14.\begin{array}{r} \ N_{\left( - \frac{1}{2} \right)}^{\pm} = \frac{1}{4}\left( ( - 1)^{\frac{1}{2}} - 2\left( - \frac{1}{2} \right)( - 1)^{\frac{1}{2}} + 1 \right) = \frac{1}{4}. \end{array}
(1.6.198)
Figure
Figure 1.6.18 – Nk±=14((1+2k)(cos⁡(kπ)+isin(kπ))+1)N_{k}^{\pm} = \frac{1}{4}\left( (1 + 2k)\left( \cos(k\pi) + isin(k\pi) \right) + 1 \right) for k∈ℝ=[−2.5,1.5]k\in \mathbb{R} =\lbrack - 2.5,1.5\rbrack.

So, the center of sym­me­try at N(−12)±=14N_{\left( - \frac{1}{2} \right)}^{\pm} = \frac{1}{4}, for the real wave­form of the func­tion Nk±N_{k}^{\pm}, is due to an­a­lyt­ic con­tin­u­a­tion, and this is a process that we will ex­plore thor­ough­ly for other func­tions in the next book. The reg­u­lar­iza­tion of the Nk±N_{k}^{\pm} series, makes it pos­si­ble to con­sid­er the reg­u­lar­iza­tion of the entire outer-prod­uct array from Eq. 1.6.181, which was formed by the al­ter­nat­ing nat­ur­al number series and the al­ter­nat­ing har­mon­ic series

SH=(∑k=1∞(−1)k+1k)⊗(∑k=1∞(−1)k+1k).\begin{array}{r} S_{H} = \left( \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k} \right) \otimes \left( \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}}{k} \right). \end{array}
(1.6.199)

So, we can con­sid­er the reg­u­lar­iza­tion of the SHS_{H} outer-prod­uct array as the sum­ma­tion of the reg­u­lar­ized al­ter­nat­ing in­verse har­mon­ic/nat­ur­al number series

∑k=1∞(−1)k+1k=14,\begin{array}{r} \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k = \frac{1}{4}}, \end{array}
(1.6.200)

mul­ti­plied by each value in the al­ter­nat­ing har­mon­ic series as we pro­ceed right­ward by column, re­sult­ing in the series, which con­verges to ln⁡(2)4\frac{\ln(2)}{4}

∑k=1∞(−1)k+14k=ln⁡(2)4.\begin{array}{r} \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}}{4k} = \frac{\ln(2)}{4}. \end{array}
(1.6.201)

This leads to the fol­low­ing de­f­i­n­i­tion for the outer-prod­uct reg­u­lar­iza­tion:

De­f­i­n­i­tion (1.6.202) – The Reg­u­lar­ized In­fi­nite Outer-Prod­uct Sum­ma­tion is a sum of el­e­ments, from two or more in­fi­nite series, that con­verges to a value or a func­tion, where at least one of the series that forms this outer-prod­uct is di­ver­gent but reg­u­lar­iz­able, and the other series(s) may or may not be an in­de­pen­dent­ly con­ver­gent series.

Let ℛ(A)\mathcal{R}(A), denote a reg­u­lar­ized sum­ma­tion for an in­fi­nite series of el­e­ments, aka_{k},

ℛ(A)=ℛ(∑k=0∞ak),\mathcal{R}(A)= \mathcal{R}\left( \sum_{k = 0}^{\infty}a_{k} \right),

which can form an outer prod­uct with an­oth­er in­fi­nite series or func­tion ex­pan­sion, BB, that may or may not con­verge in­de­pen­dent­ly. For an in­de­pen­dent­ly con­ver­gent series, BB, the outer prod­uct with ℛ(A)\mathcal{R}(A), is

ℛ(A)⊗B=ℛ(∑k=0∞ak)⊗(∑k=0∞bk),\mathcal{R}(A) \otimes B= \mathcal{R}\left( \sum_{k = 0}^{\infty}a_{k} \right) \otimes \left( \sum_{k = 0}^{\infty}b_{k} \right),

having the fol­low­ing Reg­u­lar­ized In­fi­nite Outer-Prod­uct Sum­ma­tion

∑ℛ(A)⊗B,\sum_{}^{}{\mathcal{R}(A) \otimes B},

and for a di­ver­gent but reg­u­lar­iz­able series BB, this forms the outer prod­uct with ℛ(A)\mathcal{R}(A),

ℛ(A)⊗ℛ(B)=ℛ(∑k=0∞ak)⊗ℛ(∑k=0∞bk),\mathcal{R}(A)\otimes \mathcal{R}(B)= \mathcal{R}\left( \sum_{k = 0}^{\infty}a_{k} \right)\otimes \mathcal{R}\left( \sum_{k = 0}^{\infty}b_{k} \right),

having the Reg­u­lar­ized In­fi­nite Outer-Prod­uct Sum­ma­tion

∑ℛ(A)⊗ℛ(B).\sum_{}^{}{\mathcal{R}(A)\otimes \mathcal{R}(B).}

By having a de­f­i­n­i­tion for the reg­u­lar­ized in­fi­nite outer-prod­uct sum­ma­tion, we can ex­press the pre­vi­ous ex­am­ple for the two al­ter­nat­ing har­mon­ic series as

ℛ(∑k=1∞(−1)k+1k)⊗(∑k=1∞(−1)k+1k),\begin{array}{r} \mathcal{R}\left( \sum_{k = 1}^{\infty}{( - 1)^{k + 1}k} \right) \otimes \left( \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}}{k} \right), \end{array}
(1.6.203)

which has the sum­ma­tion

∑(ℛ(∑H−k−)⊗∑Hk−)=∑Hk−4=ln⁡(2)4\begin{array}{r} \sum_{}^{}{\left( \mathcal{R}\left( \sum_{}^{}H_{- k}^{-} \right) \otimes \sum_{}^{}H_{k}^{-} \right) = \sum_{}^{}{\frac{H_{k}^{-}}{4} =}}\frac{\ln(2)}{4} \end{array}
(1.6.204)

To demon­strate a reg­u­lar­ized in­fi­nite outer-prod­uct sum­ma­tion that con­verges to a func­tion, we can use a vari­able z∈ℂz\in \mathbb{C}, to con­sid­er the outer-prod­uct

ℛ(∑k=1∞(−1)k+1zk)⊗(∑k=1∞(−1)k+1(z−1)kk),\begin{array}{r} \mathcal{R}\left( \sum_{k = 1}^{\infty}{( - 1)^{k + 1}zk} \right) \otimes \left( \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}(z - 1)^{k}}{k} \right), \end{array}
(1.6.205)

where the right­hand sum will con­verge to ln⁡(z)\ln(z), for |z−1|<1∧z=2|z - 1| < 1 \land z = 2, and the reg­u­lar­ized in­fi­nite outer-prod­uct sum­ma­tion will con­verge to the fol­low­ing func­tion

∑(ℛ(∑k=1∞(−1)k+1zk)⊗(∑k=1∞(−1)k+1(z−1)kk))=ℛ(z4)×ln⁡(z)=z4ln⁡(z).\begin{array}{r} \sum_{}^{}\left( \mathcal{R}\left( \sum_{k = 1}^{\infty}{( - 1)^{k + 1}zk} \right) \otimes \left( \sum_{k = 1}^{\infty}\frac{( - 1)^{k + 1}(z - 1)^{k}}{k} \right) \right)= \mathcal{R}\left( \frac{z}{4} \right) \times \ln(z) = \frac{z}{4}\ln(z). \end{array}
(1.6.206)

The Basel prob­lem stumped many math­e­mati­cians after first being pro­posed by Pietro Men­goli in 1650, where the prob­lem was to pre­cise­ly de­ter­mine the sum­ma­tion of the squared rec­i­p­ro­cals of the nat­ur­al num­bers, de­fined as the in­fi­nite sum

∑k=1∞1k2.\begin{array}{r} \sum_{k = 1}^{\infty}\frac{1}{k^{2}}. \end{array}
(1.6.207)

It was ul­ti­mate­ly solved by Leon­hard Euler1 in 1734, when he found the sum equaled π2/6\pi^{2}/6; a sur­pris­ing result, given the pres­ence of π\pi. The os­cil­la­tions that we have noted in our par­tial sum­ma­tion of the in­verse al­ter­nat­ing har­mon­ic num­bers, hint at the pres­ence of cir­cu­lar phe­nom­e­non in this series as well. Taking a closer look at the equa­tion,

Nk±=14((−1)k+1−2k(−1)k+1),\begin{array}{r} N_{k}^{\pm} = \frac{1}{4}\left( ( - 1)^{k + 1} - 2k( - 1)^{k} + 1 \right), \end{array}
(1.6.208)

we have that, (−1)k+1( - 1)^{k + 1}& 2k(−1)k2k( - 1)^{k} are al­ge­bra­ic forms of the fol­low­ing com­plex ex­po­nen­tial func­tions

(−1)k+1=ei(kπ+π),\begin{array}{r} ( - 1)^{k + 1} = e^{i(k\pi + \pi)}, \end{array}
(1.6.209)
2k(−1)k=2keikπ,\begin{array}{r} 2k( - 1)^{k} = 2ke^{ik\pi}, \end{array}
(1.6.210)

lead­ing to the equiv­a­lent ex­pres­sion,

Nk±=1+2keikπ+ei(kπ+π)4,\begin{array}{r} N_{k}^{\pm} = \frac{1 + 2ke^{ik\pi} + e^{i(k\pi + \pi)}}{4}, \end{array}
(1.6.211)

and can be fur­ther sim­pli­fied to

Nk±=1−(2k+1)eikπ4.\begin{array}{r} N_{k}^{\pm} = \frac{1 - (2k + 1)e^{ik\pi}}{4}. \end{array}
(1.6.212)

Like the Basel sum­ma­tion, we find the pres­ence of not only π\pi lurk­ing in the in­verse al­ter­nat­ing har­mon­ic num­bers, but also the com­plex ex­po­nen­tial func­tion. The har­mon­ic series, we have briefly dis­cussed in this sec­tion, plays an im­por­tant role in analy­sis, pro­vid­ing the ground­work for the next book, ul­ti­mate­ly taking us to the heart of prime number theory.

  1. Euler, Leonard, "De summis se­rierum rec­i­p­ro­carum", Com­men­tarii Acad­e­mi­ae Sci­en­tiarum Pet­ro­pol­i­tanae, Volume 7:123-134, 1735