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EDITs
- extension to alternating sums (6)
- extension to general sums with parameter $x$ (8)
- extension to general sums with two parameters (11)

Extended post

The relation

$$\sum _{k=1}^{n } \left(\frac{H_k^{(p)}}{k^q}+\frac{H_k^{(q)}}{k^p}\right)=H_n^{(p)} H_n^{(q)}+H_n^{(p+q)}\tag{1}$$

where $H_k^{(p)} = \sum_{j=1}^k j^{-p}$, is a nice and important symmetry relation with various applications. Notice the finite upper index $n$. The proof is not hard.

Corollary 1

If $q = p$ the relation simplifies to

$$\sum _{k=1}^n \frac{H_k^{(p)}}{k^p}= \frac{1}{2}\left( (H_n^{(p)})^2+H_n^{(2p)} \right)\tag{2}$$

Corollary 2

In the limit $n \to \infty$ (1) turns into

$$\sum _{k=1}^{\infty } \left( \frac{H_k^{(p)}}{k^q}+\frac{H_k^{(q)}}{k^p}\right)=\zeta(p)\zeta(q)+ \zeta(p+q)\tag{3}$$

where

$$\zeta(s) = \lim_{n\to \infty } \, H_n^{(s)}=\sum_{k=1}^\infty k^{-s}\tag{3a}$$

is the Riemann zeta function.

For $q=p$ formula (3) gives the analogue to (2)

$$\sum _{k=1}^{\infty } \frac{H_k^{(p)}}{k^p}=\frac{1}{2}\left(\zeta(p)^2+ \zeta(2p)\right)\tag{4}$$

This formula was also given in [1], and there it was claimed that it is valid for real $p$.

Alternating sums

A similar relation can be derived for alternating sums.

Define

$$A_n^{(p)}= \sum_{k=1}^n (-1)^k \frac{1}{k^p}\tag{5}$$

then the basic relation is

$$\sum _{k=1}^{n} (-1)^k \frac{H_k^{(p)}}{k^q} +\sum _{k=1}^{n} \frac{A_k^{(q)}}{k^p} = H_n^{(p)} A_n^{(q)}+A_n^{(p+q)}\tag{6}$$

From (6) other corollaries and formulas can be derived in a manner similar to the one for the non-alternating sums described above.

Generalized relation with parameter $x$

It can be shown with similar methods as before that a more general relation holds from which the two formulas discussed before are special cases.

Let

$$H_n^{(p)}(x) = \sum_{k=1}^n \frac{x^k}{k^p}\tag{7}$$

$H_n^{(p)}$ is understood as $H_n^{(p)}(+1)$ and we have $H_n^{(p)}(-1)=A_n^{(p)}$ (c.f. (5)).

Then the relation is

$$\sum_{k=1}^n \left(x^k \frac{H_k^ {(p)}(1)}{k^q} + \frac{H_k^{(q)}(x)}{k^p}\right)=H_n^{(p)}(1)H_n^{(q)}(x)+ H_n^{(p+q)}(x)\tag{8}$$

Notice that this relation has lost the symmetry.

For $x=1$ and $x=-1$ we recover the relations (1) and (6), respectively.

In the limit $n\to\infty$ we encounter on the right hand side the polylog function

$$\lim_{n\to\infty}\, H_n^{(p)}(x)=\sum_{k=1}^\infty \frac{x^k}{k^p} = Li_p(x)\tag{9}$$

so that we have

$$\sum_{k=1}^{\infty} \left(x^k \frac{H_k^ {(p)}(1)}{k^q} + \frac{H_k^{(q)}(x)}{k^p}\right)=Li_p(1)Li_q(x)+ Li_{p+q}(x)\tag{10}$$

Note "added in proof"

The special case $x=-1$ of relation (10) can be uncovered (it is there more or less conceiled) from [2], page 33, where it is called "shuffle relation". We now have here an elementary proof of a generalization it.

Generalized relation with two parameters $x$ and $y$

We can even go one final step further and write down this two-parametric symmetry relation

$$\sum_{k=1}^n \frac{x^k}{k^q}\sum_{m=1}^k \frac{y^m}{m^p} + \sum_{k=1}^n \frac{y^k}{k^p}\sum_{m=1}^k \frac{x^m}{m^q}=\left( \sum_{k=1}^n \frac{x^k}{k^q}\right)\left( \sum_{k=1}^n \frac{y^k}{k^p}\right)+ \sum_{k=1}^n \frac{(x y)^k}{k^{p+q}}\tag{11}$$

This relation includes the various cases of alternating and non-alternating sums for apropriate choices of $x$ and $y$.

In the limit $n\to\infty$ we obtain

$$\sum_{k=1}^\infty \frac{x^k}{k^q}\sum_{m=1}^k \frac{y^m}{m^p} + \sum_{k=1}^\infty \frac{y^k}{k^p}\sum_{m=1}^k \frac{x^m}{m^q}=\text{Li}_q(x) \text{Li}_p(y)+ \text{Li}_{p+q}(x y)\tag{12}$$

The zeta functions of (3) have been generalized to polylog functions.

Question

Prove (1), (6), (8), (11) and (12)

References

[1] Identities For Generalized Harmonic Number

[2] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf

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  • $\begingroup$ If I may ask a question. You state the proof of (1) and (6) are not too hard. Why then are you looking for proofs of these? Is it because you are hoping to see some alternative proofs for these results? $\endgroup$ – omegadot Jan 4 '18 at 23:24
  • $\begingroup$ Yes, (1) to see more elegant proofs than mine, and (2) to encourage people to join in and (3) perhaps even somebody tells me that these nice relations are well known. Quite normal and common motivations here, don't you think? And: thanks for your answer. In fact my proof was even shorter (just changing the order of summation). $\endgroup$ – Dr. Wolfgang Hintze Jan 5 '18 at 16:26
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To prove (1) I will use summation by parts in the form given by $$\sum_{k = 1}^n f_k g_k = f_n G_n - \sum_{k = 1}^{n - 1} G_k (f_{k + 1} - f_k), \quad \text{where} \quad G_n = \sum_{k = 1}^n g_k.$$

Now consider the sum $$\sum_{k = 1}^n \frac{H_k^{(p)}}{k^q}.$$ Let $f_k = H_k^{(p)}$ and $g_k = 1/k^q$. So $$G_n = \sum_{k = 1}^n \frac{1}{k^q} = H^{(q)}_n,$$ and $$f_{k + 1} - f_k = H^{(p)}_{k + 1} - H^{(p)}_k = \left (H^{(p)}_k + \frac{1}{(k + 1)^p} \right ) - H^{(p)}_k = \frac{1}{(k + 1)^p},$$ where we have made use of the following result for the generalised harmonic numbers $$H^{(a)}_{n + 1} = H^{(a)}_n + \frac{1}{(n + 1)^a}.$$

On applying the summation by parts result to our sum we have $$\sum_{k = 1}^n \frac{H^{(p)}_k}{k^q} = H^{(p)}_n H^{(q)}_n - \sum_{k = 1}^{n - 1} \frac{H^{(q)}_k}{(k + 1)^p}.$$ Shifting the index is the sum appearing to the right gives \begin{align*} \sum_{k = 1}^n \frac{H^{(p)}_k}{k^q} &= H^{(p)}_n H^{(q)}_n - \sum_{k = 2}^{n} \frac{H^{(q)}_{k - 1}}{k^p}\\\ &= H^{(p)}_n H^{(q)}_n - \sum_{k = 2}^{n} \frac{1}{k^p} \left (H^{(q)}_k - \frac{1}{k^q} \right )\\ &= H^{(p)}_n H^{(q)}_n - \sum_{k = 2}^{n} \frac{H^{(q)}_k}{k^p} + \sum_{k = 2}^n \frac{1}{k^{p+q}}\\ &= H^{(p)}_n H^{(q)}_n - \sum_{k = 1}^{n} \frac{H^{(q)}_k}{k^p} + \sum_{k = 1}^n \frac{1}{k^{p+q}}\\ &= H^{(p)}_n H^{(q)}_n - \sum_{k = 1}^{n} \frac{H^{(q)}_k}{k^p} + H^{p + q)}_n \end{align*} or after arranging $$\sum_{k = 1}^n \left (\frac{H^{(p)}_k}{k^q} + \frac{H^{(q)}}{k^p} \right ) = H^{(p)}_n \cdot H^{(q)}_n + H^{(p + q)}_n,$$ as required to show.

The result given by (6) can be proved in a similar fashion as to what was done above using summation by parts.

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Proof

@omegadot has given a proof using partial summation. My proof is based on interchanging the order of summation.

Starting with $\sum_{k=1}^n \frac{H_k^{(p)}}{k^q}$, then inserting the definition $H_k^{(p)} = \sum_{i=1}^k \frac{1}{i^p}$ we arrive at a double sum in which we $\color{red}{\text{interchange the order of summation}}$ (notice the change in the ranges of the indices), $\color{violet}{\text{complete the sum (and subtract the correction)}}$, and $\color{blue}{\text{shift an index}}$ as follows

$$\sum_{k=1}^n \frac{H_k^{(p)}}{k^q}=\color{red}{\sum_{k=1}^n \sum_{i=1}^k \frac{1}{k^q}\frac{1}{i^p} = \sum_{i=1}^n \sum_{k=i}^n \frac{1}{i^p}\frac{1}{k^q}} = \left( \sum_{i=1}^n \frac{1}{i^p}\right)\color{violet}{\left(\sum_{k=1}^n \frac{1}{k^q} - \sum_{k=1}^{i-1} \frac{1}{k^q} \right)} =\left( \sum_{i=1}^n \frac{1}{i^p}\right)\left(\sum_{k=1}^n \frac{1}{k^q} \color{blue}{- \sum_{k=1}^{i} \frac{1}{k^q}+\frac{1}{i^q}} \right) =\left(\sum_{i=1}^n \frac{1}{i^p}\right)\left(\sum_{k=1}^n \frac{1}{k^q} \right) - \sum_{i=1}^n \frac{H_i^{(q)}}{i^p}+\sum_{i=1}^n \frac{1}{i^q i^p} = H_n^{(p)} H_n^{(q)} + H_n^{(p+q)} - \sum_{i=1}^n \frac{H_i^{(q)}}{i^p} \tag{s1}$$

Rearranging gives (1). QED.

The proof can easily be extended to (6).

Generalization and visualization

The general version of the symmetry relation is

$$\sum_{k=1}^n a_k \sum_{i=1}^k b_i + \sum_{i=1}^n b_i \sum_{k=1}^i a_k= (\sum_{k=1}^n a_k)(\sum_{i=1}^n b_i)+ \sum_{k=1}^n a_k b_k\tag{s2}$$

The basic interchange rule

$$\sum_{k=1}^n a_k \sum_{i=1}^k b_i=\sum_{i=1}^n b_k \sum_{k=i}^n a_k\tag{s3} $$

can be illustrated as follows.

The left hand side of (s3) can be written as

$$\begin{array}\\ &a_1(b_1)\\ +&a_2(b_1+b_2)\\ +&a_3(b_1+b_2+b_3)\\ +&...\\ +&a_n(b_1+b_2+b_3+...+b_n) \end{array}$$

Summing vertically gives

$$\begin{array}\\ &b_1(a_1+a_2+a_3+...+a_n)\\ +&b_2(\;\;\;\; +\;a_2+ a_3+...+a_n)\\ +&b_3(\;\;\;\;\;\;\;\;\;\;\;\;+\; a_3+...+a_n)\\ +&...\\ +&b_n(\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+ a_n) \end{array}$$

which is the right hand side of (s3). done.

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