Evaluation of convergent series by using finite parts

1. Introduction

The aim of this article is to present a method for the summation of convergent series by using the finite part of divergent series in the intermediate steps. Using divergent series to find the result of convergent processes is a very old practice as can be seen very clearly by reading chapter 1 of Hardy's book [1] or by consulting classic texts on series such as Bromwich's [2]; a great variety of summability methods have been employed throughout the years [1]. This article differs from older methods, however, in that we employ Hadamard finite part limits in our process. Finite part limits are explained in Sections 3 and 4, and the many examples given should convince the reader that the extraction of the finite part of a series that diverges to infinity is a regularization procedure, not a summability method. Hadamard finite parts of limits and integrals have been employed mainly in the areas of partial differential equations, since they were first defined by Hadamard [3] to find fundamental solutions of hyperbolic partial differential equations, and in the theory of distributions and generalized functions [4, 5], where they play a rather important part, mostly because of their relationship with pseudofunctions.

In Section 5, we find the sum of the convergent series by finding the finite part of the sums of the two divergent series and substracting those results. Next, in Section 6, we consider a more complicated example, namely the convergent series where we employ standard ideas such as Abel's summation formula and Stirling's asymptotic approximation to find the finite part of the divergent series encountered.
Sections 7 and 8 contain a more systematic treatment of a class of convergent series that generalize (1) and (3). The main tool of our analysis is the formula for the finite part of the divergent series \(\sum_{n=0}^{\infty}\left( n+a\right) ^{k}\ln\left( n+a\right) ,\) where \(H_{k}=\sum_{j=1}^{k}\left( 1/j\right) \) are the harmonic numbers.
We are then able to find the sum of certain convergent series involving values of the Hurwitz zeta function, namely, those of the form for \(\left\vert b\right\vert \leq\left\vert a\right\vert ,\) \(b\neq-a,\) \(k\in\mathbb{N}.\) Series of this family have received a lot of attention since the times of Euler, who gave the interesting sums and where \(\gamma\) is Euler's constant. Srivastava [6] considers many of these series in an article that contains a very detailed historical account of the contributions of several authors. Recently, Blagouchine [7] and Coppo [8] have given general formulas for the series for all \(k,\) that generalize not only (6) but also the nice Surnayanarayana's formula [9] Interestingly, Surnayanarayana also considers divergent but Ces\`{a}ro summable sums in [9], as the formula that can also be found by employing exterior Euler summability [10] and is also ammenable of our analysis. Notice that the series (5) diverges to infinity if \(b=-a.\) Nevertheless, after reading this article the reader will be able to see that

Preliminaries

In this section, we fix the notation employed and recall several useful well known facts; see [11] and [12] for details.
We shall always consider the principal branches of powers and logarithms, \(z^{\alpha}\) and \(\ln z,\) defined for \(z\in\mathbb{C}\setminus(-\infty,0].\) If \(a\in\mathbb{C}\setminus(-\infty,0],\) the Hurwicz zeta function \(\zeta\left( s,a\right) \)\ is the analytic continuation to \(s\in\mathbb{C}\setminus \left\{ 1\right\} \) of the function The zeta function \(\zeta\left( s,a\right) \)\ has a simple pole at \(s=1,\) with residue \(1.\) If \(a=1\) it reduces to the Riemman zeta function, \(\zeta\left( s,1\right) =\zeta\left( s\right) .\)\ We shall also employ the digamma function \(\psi\left( a\right) =\Gamma^{\prime}\left( a\right) /\Gamma\left( a\right) ,\) the logarithmic derivative of the gamma function, whose Mittag-Leffer expansion is the following where \(\gamma\) is Euler's constant. The digamma function is actually the finite part -- explained in the next section -- of the negative of the Hurwitz zeta function, \(-\zeta\left( s,a\right) ,\)\ at \(s=1,\) that is for \(a\in\mathbb{C}\setminus(-\infty,0].\) Notice also that It is easy to see that for \(n\geq1,\) and this yields the Taylor expansion which is an absolutely convergent series for \(\left\vert \omega\right\vert < \left\vert a\right\vert .\) The series is divergent for \(\left\vert \omega\right\vert =\left\vert a\right\vert \) but it is Ces\`{a}ro summable in the this circle except for \(\omega\neq-a.\)
Sometimes, we shall need to use Abel's summation formula [1, Chp. 1], where \(A_{n}=\sum_{j=1}^{n}a_{j}.\) We shall also employ Stirling's asymptotic formula [12], as \(z\rightarrow\infty\) in the region \(-\pi<\arg z<\pi.\)

3. The finite part of limits

Let us now recall the notion of the finite part of a limit [13, Section 2.4]. Let \(X\) be a topological space, and let \(x_{0}\in X.\) Suppose \(\mathfrak{F},\) the basic functions, is a family of strictly positive functions defined for \(x\in V\setminus\left\{ x_{0}\right\} ,\) where \(V\) is a neighborhood of \(x_{0},\) such that all of them tend to infinity at \(x_{0}\) and such that, given two different elements \(f_{1},f_{2}\in\mathfrak{F}\), then \(\lim_{x\rightarrow x_{0}}f_{1}\left( x\right) /f_{2}\left( x\right) \) is either \(0\) or \(\infty.\)

Definition 1. \label{Def FP}Let \(G\left( x\right) \) be a function defined for \(x\in V\setminus\left\{ x_{0}\right\} \) with \(\lim_{x\rightarrow x_{0}}G\left( x\right) =\infty.\) The finite part of the limit of \(G\left( x\right) \) as \(x\rightarrow x_{0}\) with respect to \(\mathfrak{F}\) exists and equals \(A\) if we can write\footnote{Such a decomposition, if it exists, is unique since any finite number of elements of \(\mathfrak{F}\) has to be linearly independent.} \(G\left( x\right) =G_{1}\left( x\right) +G_{2}\left( x\right) ,\) where \(G_{1},\) the infinite part, is a linear combination of the basic functions and where \(G_{2},\)\ the finite part, has the property that the limit \(A=\lim_{x\rightarrow x_{0}}G_{2}\left( x\right) \) exists. We then employ the notation

We shall be interested in the Hadamard finite parts at infinity in this article. It is the case when \(\mathfrak{F}\) is the family of powers and logarithms.

Definition 2. The Hadamard finite part limit corresponds to the case when \(x_{0}=0\) and \(\mathfrak{F}\) is the family of functions \(x^{-\alpha}\left\vert \ln x\right\vert ^{\beta},\) where \(\alpha>0\) and \(\beta\geq0\) or where \(\alpha=0\) and \(\beta>0\) or when \(x_{0}=\infty\) and \(\mathfrak{F}\) is the family of functions \(x^{\alpha}\left\vert \ln x\right\vert ^{\beta},\) where \(\alpha>0\) and \(\beta\geq0\) or where \(\alpha=0\) and \(\beta>0.\) We then use the simpler notations \(\mathrm{F.p.}\lim_{x\rightarrow0^{+}}G\left( x\right) \) or \(\mathrm{F.p.}\lim_{x\rightarrow\infty}G\left( x\right) .\)

Applying the notion of finite part of limits we can define the finite part of integrals at singularities [13, Section 2.4], but in this article, we will be interested in the finite part of series,

Observe that the finite part of a series gives a finite sum to a series that diverges to infinity; it does not work for oscillatory series. This makes the finite part process very different from summability methods like Abel summability of Ces\`{a}ro summability. The finite part process is more of a regularization procedure than a summability method [13, Section 2.4].

Needless to say, many of the usual operations that are valid for convergent series may not work for the finite parts. For example, \(\mathrm{F.p.} \sum_{n=n_{0}+1}^{\infty}a_{n-1}\) does not have to be equal to \(\mathrm{F.p.} \sum_{n=n_{0}}^{\infty}a_{n}.\)

The notion of finite part integrals and its name were introduced by Hadamard [3], who used them in his study of fundamental solutions of partial differential equations. The finite part method has been a very important part of the theory of distributions since its beginnings [5] since it is closely related to the notion of pseudofunction.

4. Some finite part limits

In this section, for future reference, we will give the formulas for several useful finite part limits. Our definition says that \(\mathrm{F.p.}\lim_{N\rightarrow\infty}N^{\beta}=0\) for all \(\beta\in\mathbb{R}.\) While it is true that for all \(a\in\mathbb{C},\) we have In a similar fashion, while \(\mathrm{F.p.}\lim_{N\rightarrow\infty}N^{\beta }\ln N=0\) for all \(\beta\in\mathbb{R},\) by definition, when we replace \(N\) by \(N+a\) we have the ensuing formulas.

Lemma 1. If \(a\in\mathbb{C}\) and \(\beta\in\mathbb{R}\setminus \left\{ 1,2,3,\ldots\right\} \) then

If \(k=1,2,3,\ldots\) then where \(H_{k}\) is the harmonic sum

Proof. If \(k\in\left\{ 1,2,3,\ldots\right\} \) then \begin{align*} \mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+a\right) ^{k}\ln\left( N+a\right) & =\mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+a\right) ^{k}\ln N+\mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+a\right) ^{k} \ln\left( 1+\frac{a}{N}\right) \\ & =\mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+a\right) ^{k}\ln\left( 1+\frac{a}{N}\right) \,, \end{align*} while for \(N\) large \[ \left( N+a\right) ^{k}\ln\left( 1+\frac{a}{N}\right) =\left( \sum _{j=0}^{k}\binom{k}{j}N^{k-j}a^{j}\right) \sum_{n=1}^{\infty}\left( -1\right) ^{n+1}\frac{a^{n}}{nN^{n}}\,, \] so that

Observe now that \begin{eqnarray*} k\binom{k}{1}-\frac{1}{2}\binom{k}{2}+\cdots+\frac{\left( -1\right) ^{k}}{k}\binom{k}{k}=\int_{0}^{1}\frac{1-\left( 1-x\right) ^{k}}{x} \,\mathrm{d}x =\int_{0}^{1}\frac{1-t^{k}}{1-t}\,\mathrm{d}t=\int_{0}^{1}\left( 1+t+\cdots+t^{k-1}\right) \,\mathrm{d}t=H_{k}, \end{eqnarray*} which yields (25).

Next, we would like to consider the finite part limit of several finite sums. We have that \(\sum_{n=1}^{\infty}n^{-\beta}=\zeta\left( \beta\right) ,\) a convergent series if \(\Re e\,\beta>1.\) We now consider the finite part limit \(\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=1}^{N}n^{-\beta}\) for \(\Re e\,\beta\leq1;\) do we obtain \(\zeta\left( \beta\right) ?\) The answer is yes almost all the time, but not always.

Lemma 2. If \(\Re e\,\beta\in\mathbb{R}\setminus\left\{ 1,0,-1,-3,-5,\ldots\right\} \) then

If \(\beta=-k=0,1,3,5,\ldots\) then while if \(\beta=1,\) The finite part limit \(\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=1} ^{N}n^{-\beta}\) does not exist if \(\Re e\,\beta\in\left\{ 1,0,-1,-3,-5,\ldots \right\} ,\) \(\Im m\,\beta\neq0.\)

Proof. Formula (29) is trivial, while (30) follows from the well known asymptotic relation

In the other cases, we just employ the asymptotic expansion \begin{eqnarray} \zeta\left( s\right) \sim\sum_{n=1}^{N}\frac{1}{n^{s}}-\frac{N^{1-s} }{1-s}+\zeta\left( 0\right) N^{-s}\label{spf a11} -\sum_{q=0}^{\infty}\frac{\zeta\left( -2q-1\right) s\left( s+1\right) \cdots\left( s+2q\right) }{\left( 2q+1\right) !}N^{-s-2q-1},\nonumber \end{eqnarray} that comes from the Euler-Maclaurin formula.

It is interesting that the formula (28) holds for some negative integers, namely, for \(-2,\) \(-4,\) \(-6,\) \(\ldots.\) Observe also that a simple modification of the idea of finite part would make (28) hold when \(\Re e\,\beta\in\left\{ 1,0,-1,-3,-5,\ldots\right\} ,\) \(\Im m\,\beta\neq0,\) but we shall not need to consider this case in this article.

Similar formulas are obtained if we replace \(n\) by \(n+a\) in the previous lemma.

Lemma 3. Let \(a\in\mathbb{C}\setminus(-\infty,0].\) If \(\Re e\,\beta\in\mathbb{R}\setminus\left\{ 1,0,-1,-3,-5,\ldots\right\} \) then

If \(\beta=-k=0,1,3,5,\ldots\) then while if \(\beta=1,\) The finite part limit \(\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=0} ^{N}\left( n+a\right) ^{-\beta}\) does not exist in case \(\Re e\,\beta \in\{1,\) \(0,\) \(-1,\) \(-3,\) \(-5,\) \(\ldots\},\) \(\Im m\,\beta\neq0.\)

Proof. The proof is basically the same as that of the previous lemma if we use the asymptotic expansion (56) instead of (32) and the formula [14]

that corresponds to (31).

It is important to point out that while \(\zeta\left( \beta\right) \) is obtained by putting \(a=1\) in \(\zeta\left( \beta,a\right) ,\) the formulas in the Lemmas 2 and 3, are not exactly the same: the fact that we obtain \(a^{k}\) in (33) and not \(0\) as in (29) is explained because one sum is from \(n=0\) to \(N\) while the other from \(n=1\) to \(N.\)

5. A simple example

Let us consider the convergent series where \(a\) is a constant, \(a\neq0,-1,-2,\ldots.\) We can easily evaluate it by using finite parts, since we have a telescoping sum, which, when we notice that \(\mathrm{F.p.}\lim_{N\rightarrow\infty}\ln\left( N+a+1\right) =0,\) gives the finite part limit while from (34) \(\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum _{n=0}^{N}\left( n+a\right) ^{-1}=-\psi\left( a\right) .\) Hence There is also a simple alternative evaluation of \(\sigma_{0}\left( a\right) ,\) without the use of finite parts, since (13) gives \begin{align*} \int_{0}^{1}\psi\left( a+\omega\right) \,\mathrm{d}\omega & =-\gamma +\sum_{n=0}^{\infty}\left( \frac{1}{n+1}-\int_{0}^{1}\frac{1}{n+a+\omega }\,\mathrm{d}\omega\right)=-\gamma+\sum_{n=0}^{\infty}\left( \frac{1}{n+1}-\ln\left( \frac {n+a+1}{n+a}\right) \right) \\ & =\psi\left( a\right) +\sum_{n=0}^{\infty}\left( \frac{1}{n+a}-\ln\left( \frac{n+a+1}{n+a}\right) \right)\,, \end{align*} or and We also observe that from (17), If \(a=1\), we obtain the well known formula that goes back to Euler,

6. Another example

Our next task is to find the sum of the convergent series First, we approximate the sum by employing Abel's summation formula (18), \begin{eqnarray*} S_{N} =-\sum_{j=1}^{N}\ln\left( j+a\right) +N\ln\left( N+a+1\right) =\ln\left( \frac{\Gamma\left( a+1\right) }{\Gamma\left( N+a+1\right) }\right) +N\ln\left( N+a+1\right), \end{eqnarray*} and then Stirling's formula (19), to obtain and consequently If we now recall (39), we obtain Formulas (49) and (50) allow us to obtain an expression for the sum Indeed, (17) yields and \begin{align*} \int_{0}^{1}\omega\psi\left( a+\omega\right) \,\mathrm{d}\omega & =\int _{0}^{1}\omega\left\{-\gamma+\sum_{n=0}^{\infty}\left( \frac{1}{n+1}-\frac {1}{n+a+\omega}\right) \right\}\,\mathrm{d}\omega \\&=-\mathrm{F.p.}\lim_{N\rightarrow\infty}\int_{0}^{1}\sum_{n=0}^{N} \frac{\omega}{n+a+\omega}\,\mathrm{d}\omega\\ & =\mathrm{F.p.}\lim_{N\rightarrow\infty}\int_{0}^{1}\sum_{n=0}^{N} \left\{\frac{n+a}{n+a+\omega}-1\right\}\,\mathrm{d}\omega\\&=-1+\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=0}^{N}\left( n+a\right) \ln\left( \frac{n+a+1}{n+a}\right) \\ & =\ln\Gamma\left( a+1\right) +a-\ln\sqrt{2\pi}-a\ln a\,, \end{align*} so that \(\nu_{1}\left( a\right) =\sigma_{1}\left( a\right) ,\)\ that is, Notice that if \(a=1\) we obtain Surnayanarana's formula [9] while \(a=2\) yields the result [6]

7. The functions \(\varphi_{k}\left( a\right) \)

The Euler-Maclaurin formula yields the asymptotic development as \(N\rightarrow\infty.\) Differentiation of (56) gives the expansion Let us define We may evaluate \(\varphi_{0}\left( a\right) \) in two ways, one using Stirlings formula (19), \begin{align*} \varphi_{0}\left( a\right) & =\mathrm{F.p.}\lim_{N\rightarrow\infty }\left[ \ln\Gamma\left( N+a+1\right) -\ln\Gamma\left( a\right) \right] \\ & =\mathrm{F.p.}\lim_{N\rightarrow\infty}\left[ (N+a+\frac{1}{2})\ln\left( N+a+1\right) -\left( N+a+1\right) +\ln\sqrt{2\pi}-\ln\Gamma\left( a\right) \right] \\ & =\ln\sqrt{2\pi}-\ln\Gamma\left( a\right) \,, \end{align*} the other using (57) that gives \(\varphi_{0}\left( a\right) =-\zeta^{\prime}\left( 0,a\right) ,\) so that Next we consider \(\varphi_{1}\left( a\right) .\) We have from (57) that \(\varphi_{1}\left( a\right) \) is given as the finite part of the limit of as \(N\rightarrow\infty.\) Hence using the Lemma 1 we obtain More generally, we have the following formula.

Proposition 1. If \(k\geq1\) then

where \(H_{k}\) are the harmonic numbers (26).

Proof. It is clear from (57) that

where \(P_{k+1}\) is a certain polynomial of degree \(k+1.\) Let us now consider the difference \begin{align*} \varphi_{k}\left( a+1\right) -\varphi_{k}\left( a\right) & =\left( \zeta^{\prime}\left( -k,a\right) -\zeta^{\prime}\left( -k,a+1\right) \right) +\left( P_{k+1}\left( a+1\right) -P_{k+1}\left( a\right) \right) \\ & =-a^{k}\ln a+\left( P_{k+1}\left( a+1\right) -P_{k+1}\left( a\right) \right) \,. \end{align*} We also have \begin{align*} \varphi_{k}\left( a+1\right) -\varphi_{k}\left( a\right) & =\mathrm{F.p.}\lim_{N\rightarrow\infty}(\sum_{n=0}^{N}\left( n+a+1\right) ^{k}\ln\left( n+a+1\right) -\sum_{n=0}^{N}\left( n+a\right) ^{k}\ln\left( n+a\right) )\\ & =\mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+a+1\right) ^{k}\ln\left( N+a+1\right) -a^{k}\ln a\\ & =H_{k}\left( a+1\right) ^{k}-a^{k}\ln a\,, \end{align*} because of (25). Consequently, \(P_{k+1}\) is the a polynomial solution of the equation Actually it is the only polynomial solution that satisfies the initial condition since \(P_{k+1}\left( 0\right) \) is the constant term in the polynomial \(P_{k+1}\left( a\right) ,\) and (57) tell us that that constant term is precisely \(-H_{k}\zeta\left( -k\right) \) (that vanishes if \(k\) is even).
Furthermore,\ the Bernoulli polynomials \(B_{k}\left( a\right) \) satisfy \(B_{k}\left( x+1\right) -B_{k}\left( x\right) =kx^{k-1},\) \(B_{k}\left( 0\right) =B_{k}\left( 1\right) =-k\zeta\left( 1-k\right) \) [11] . Hence \(H_{k}B_{k+1}\left( a+1\right) /\left( k+1\right) \) is another polynomial solution of the initial value problem for the difference equation. Therefore Formula (62) is obtained since as is well known [11].

When \(a=1\) it is convenient to consider a variation of the function \(\varphi_{k}\left( 1\right) .\)

Proposition 2. \label{Prop. EM 2}If \(k\in\mathbb{N}\) then

Proof. Indeed, using the Lemma 1, we obtain \begin{align*} \mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=1}^{N}n^{k}\ln n & =\varphi_{k}\left( 1\right) -\mathrm{F.p.}\lim_{N\rightarrow\infty}\left( N+1\right) ^{k}\ln\left( N+1\right) \\ & =-\zeta^{\prime}\left( -k\right) -H_{k}\zeta\left( -k,2\right) -H_{k}=-\zeta^{\prime}\left( -k\right) -H_{k}\zeta\left( -k\right) \ \ \ \text{as required.} \end{align*}

8. The series \(\sum_{n=2}^{\infty}\left( -1\right) ^{n}\zeta\left(n,a\right) b^{n+k}/(n+k)\)

We now employ finite part limits to find a formula for the sum of some general convergent series.

Proposition 3. \label{Prop. se 1}Let \(a\in\mathbb{C}\setminus(-\infty,0]\) and let \(\left\vert b\right\vert \leq\left\vert a\right\vert ,\) \(b\neq-a.\) Then for \(k\in \mathbb{N}\)

Proof. Indeed, from (17), we obtain \begin{eqnarray*} \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\zeta\left( n,a\right) b^{n+k}}{n+k}&=&\int_{0}^{b}\left(-\psi\left( a\right) +\psi\left( a+\omega\right) \right) \omega^{k}\mathrm{d}\omega\\&=&\frac{-\psi\left( a\right)b^{k+1}}{k+1}+\int_{0}^{b}\left( -\gamma\omega^{k}+\sum_{n=0}^{\infty}\left( \frac{\omega^{k}}{n+1} -\frac{\omega^{k}}{n+a+\omega}\right) \right) \mathrm{d}\omega\\ &=&\frac{-\psi\left( a\right) b^{k+1}} {k+1}-\mathrm{F.p.}\lim_{N\rightarrow\infty}\sum_{n=0}^{N}\int_{0}^{b} \frac{\omega^{k}}{n+a+\omega}\,\mathrm{d}\omega\,. \end{eqnarray*} Observe now that

which yields Therefore If we notice now that and substitute, we obtain (69).

The cases, when \(b=\pm1\) are worth recording. Let us observe that to obtain and If we put \(a=1\) in (75), we obtain from where one may recover the formula for \(\nu_{k}\) given by Coppo [8], who rewrote the formula given earlier by Blagouchine [7]. On the other hand, setting \(a=2\) in (76) gives The case \(k=0,\) goes back to Euler, while the case \(k=1\) can be found in Srivastava [6] We would also like to point out that by adding and substracting (75) and (76), we can derive formulas for series involving only even or only odd arguments of the Hurwitz zeta function. For example and

Acknowledgments

The authors would like to express their thanks to the referees for their useful remarks and encouragements.

Authorcontributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflictofinterests

The author declares no conflict of interest.