La forma cerrada para la armónica de la aproximación de la suma de la $\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2$

Question

Pregunta

Hay una forma cerrada de este armónico aproximación de la suma de la

$$s=\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2\tag{1}$$

La notación es estándar.

La motivación

Esta pregunta se refiere a un campo que fue tratado con frecuencia por muchos contribuidores aquí donde, en general, un adecuado funcionamiento de la diferencia de un sumando un su aproximación asintótica se resume, y sobre todo que es la pregunta de si existe una forma cerrada de esta cantidad en los términos de la norma de "constantes". La aproximación así como la funcional debe ser elegido tal que el resultado de la suma infinita en convergentes. Hay muchos ejemplos de tales "aproximación sumas".

Recientemente [1] he publicado una pregunta relativa a la aproximación de la suma de la

$$\sum _{k=1}^{\infty } \left(H_k-(\log(k) + \gamma)\right){}^2\tag{2}$$

Esto llevó a que el problema de una suma que contiene una infrecuente $\log$-factor en lugar de la común polinomio en el índice.

En un intento de deshacerse de la $\log$, pero todavía conservan un no problema trivial pido aquí para encontrar una forma cerrada para $s$ definido en (1).

Una generalización natural es la generación de la función

$$s_{m,p}(x)=\sum _{k=1}^{\infty } x^k \left(H_k^{(m)}-\zeta (m)\right){}^p\tag{3}$$

A mi conocimiento, sobre la aproximación de las sumas que contiene modificado armónica de los números no han sido investigados aquí.

Es el valor numérico

$$N(s_2) = 0.900362625200937377409205241520956358081230891307664$$

Solución intento

Consideramos que las sumas parciales y escribir $s = \sigma_1+\sigma_2+\sigma_3$ donde

$$\sigma_1 = \sum _{k=1}^{n } \left(H_k^{(2)}\right){}^2\tag{4a}$$ $$\sigma_2 =-2 \zeta(2) \sum _{k=1}^{n } \left(H_k^{(2)}\right){}^2\tag{4b}$$ $$\sigma_3 = n \zeta(2)^2\tag{4c}$$

La suma de $\sigma_2$ se puede calcular:

$$\sum _{k=1}^n H_k^{(2)}=(n+1) H_{n+1}^{(2)}-H_{n+1}\tag{5}$$

así que

$$\sigma_2 =-2 \zeta(2)\left( (n+1) H_{n+1}^{(2)}-H_{n+1}\right)\tag{6}$$

y nos quedamos con $\sigma_1$ que puede ser fácilmente demostrado por el lector a reducir para el cálculo de cualquiera de las

$$h_2 = \sum _{k=1}^n \frac{H_k}{k^2}\tag{7}$$

o

$$h_4 = \sum _{k=1}^n \frac{H_{k}^{(2)}}{k}\tag{8}$$

Estas sumas se han discutido (y christianed) anteriormente en [2] donde también esta relación ha sido derivados

$$h_2+h_4 =H_n H_{n}^{(2)} + H_{n}^{(3)}\tag{9}$$

lo que significa que el conocimiento de una de estas funciones es suffient.

Sería bonito ver a una forma explícita para el (finito) de las sumas $h$ en términos de los elementos del conjunto que contiene (modificado) armónica de los números y de los escalares.

Pero para que la tarea actual sólo el límite de índice debe ser determinada.

Referencias

[1] la Generación de función para armónica número de veces de registro ($H_k log(k)$)

[2] la Suma de las potencias de Armónica de los Números

Answer 1

Por sumación por partes $$\begin{eqnarray*} -\sum_{k\geq 1}1\cdot\left(\zeta(2)-H_k^{(2)}\right)^2 &=& \sum_{n\geq 1}\frac{n}{(n+1)^4}+\frac{2n (H_n^{(2)}-\zeta(2))}{(n+1)^2}\\&=&\zeta(3)-\zeta(4)+2\sum_{n\geq 1}\frac{H_n^{(2)}-\zeta(2)}{n+1}-2\sum_{n\geq 1}\frac{H_n^{(2)}-\zeta(2)}{(n+1)^2}\\&=&\zeta(3)-\zeta(4)+2\zeta(2)-4\zeta(3)+\frac{7\pi^4-60\pi^2}{180}\end{eqnarray*}$$ (el último de la identidad de la siguiente manera estándar de Euler sumas) y simplificando obtenemos $$\boxed{ \sum_{n\geq 1}\left(\zeta(2)-H_k^{(2)}\right)^2 = \color{red}{3\,\zeta(3)-\zeta(2)^2.}}$$ Del mismo modo, $$\begin{eqnarray*}\sum_{k\geq 1}\left(\zeta(m)-H_k^{(m)}\right)^2 &=& \sum_{k\geq 1}\left[2\,\zeta(m)-H_{k}^{(m)}-H_{k+1}^{(m)}\right]\frac{k}{(k+1)^m}\\&=&\sum_{k\geq 1}\left[2\,\zeta(m)-2H_{k}^{(m)}-\frac{1}{(k+1)^m}\right]\cdot\left[\frac{1}{(k+1)^{m-1}}-\frac{1}{(k+1)^m}\right]\end{eqnarray*}$$ se reduce a calcular algunos valores de $\zeta$, recordando $\sum_{k\geq 1}\frac{H_k^{(m)}}{k^m}=\frac{\zeta(2m)+\zeta(m)^2}{2} $ (simetría) y abordar $\sum_{k\geq 1}\frac{H_k^{(m)}}{k^{m-1}}$ o $\sum_{k\geq 1}\frac{H_k^{(m-1)}}{k^m}$. Flajolet y Salvy esquema de manera eficiente a través de los residuos. Los primeros casos que se $$ \sum_{k\geq 1}\frac{H_k}{k^2}=2\,\zeta(3),\qquad \sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} = 3\,\zeta(2)\,\zeta(3)-\frac{9}{2}\,\zeta(5) $$ $$ \sum_{k\geq 1}\frac{H_k^{(3)}}{k^4} = -10\,\zeta(2)\,\zeta(5)+18\,\zeta(7). $$

Answer 2

EDITAR 03.01.18

• la alternancia de suma $s_{m,2}(-1)$ calculado para la básico de la alternancia de suma $S^{+-}_{m,m} = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(m)}}{k^m}$

• breve discusión de $S^{+-}_{m,m}$. Formas cerradas están disponibles para $m=1$ $m=2 in terms of $\zeta$-values, $\log(2)$, and Li. For $m \ge 3$ reducibilidad queda por aclarar.

• asintótica comportamiento de $s_{m,2}(1)$ numéricamente determinado

Post Original

Inspirada en la primera parte de la solución de Jack D'Aurizio puedo mostrar aquí que la suma

$$s_{m}=s_{m,2}(1) =\sum _{k=1}^{\infty } \left(\zeta (m)-H_k^{(m)}\right){}^2\tag{1}$$

has a closed form not only for $m=2$ but generally for $m\ge 3$. Furthermore, this closed form is composed solely of zeta-values, i.e. of values of Riemann's $\zeta$ function at positive integers $\ge 2$.

Result

Here are the first few expressions in the format $\{m, s_m\}$

$$ \begin{array}{c} \left\{2,- \zeta (2)^2+3\zeta (3)\right\} \\ \left\{3,-\zeta (3)^2+\pi ^2 \zeta (3)-10 \zeta (5)\right\} \\ \left\{4,-\frac{1}{3} 10 \pi ^2 \zeta (5)+35 \zeta (7)-\frac{\pi ^8}{8100}\right\} \\ \left\{5,-\zeta (5)^2+\frac{\pi ^4 \zeta (5)}{9}+\frac{35 \pi ^2 \zeta (7)}{3}-126 \zeta (9)\right\} \\ \left\{6,-\frac{1}{15} 7 \pi ^4 \zeta (7)-42 \pi ^2 \zeta (9)+462 \zeta (11)-\frac{\pi ^{12}}{893025}\right\} \\ \left\{7,-\zeta (7)^2+\frac{2 \pi ^6 \zeta (7)}{135}+\frac{28 \pi ^4 \zeta (9)}{15}+154 \pi ^2 \zeta (11)-1716 \zeta (13)\right\} \\ \left\{8,-\frac{1}{105} 8 \pi ^6 \zeta (9)-\frac{22 \pi ^4 \zeta (11)}{3}-572 \pi ^2 \zeta (13)+6435 \zeta (15)-\frac{\pi ^{16}}{89302500}\right\} \\ \left\{9,-\zeta (9)^2+\frac{\pi ^8 \zeta (9)}{525}+\frac{22 \pi ^6 \zeta (11)}{63}+\frac{143 \pi ^4 \zeta (13)}{5}+2145 \pi ^2 \zeta (15)-24310 \zeta (17)\right\} \\ \left\{10,-\frac{1}{945} 11 \pi ^8 \zeta (11)-\frac{286 \pi ^6 \zeta (13)}{189}-\frac{1001 \pi ^4 \zeta (15)}{9}-\frac{24310 \pi ^2 \zeta (17)}{3}+92378 \zeta (19)-\frac{\pi ^{20}}{8752538025}\right\} \\ \end{array}\etiqueta{2a} $$

Or, in a "canonical" form, in which all powers of $\pi$ are replaced by corresponding $\zeta$-functions

$$ \begin{array}{cc} 2 & - \zeta (2)^2+3\zeta (3)\\ 3 & -\zeta (3)^2+6 \zeta (2) \zeta (3)-10 \zeta (5) \\ 4 & -20 \zeta (2) \zeta (5)+35 \zeta (7)-\frac{7 \zeta (8)}{6} \\ 5 & -\zeta (5)^2+10 \zeta (4) \zeta (5)+70 \zeta (2) \zeta (7)-126 \zeta (9) \\ 6 & -42 \zeta (4) \zeta (7)-252 \zeta (2) \zeta (9)+462 \zeta (11)-\frac{715 \zeta (12)}{691} \\ 7 & -\zeta (7)^2+14 \zeta (6) \zeta (7)+168 \zeta (4) \zeta (9)+924 \zeta (2) \zeta (11)-1716 \zeta (13) \\ 8 & -72 \zeta (6) \zeta (9)-660 \zeta (4) \zeta (11)-3432 \zeta (2) \zeta (13)+6435 \zeta (15)-\frac{7293 \zeta (16)}{7234} \\ 9 & -\zeta (9)^2+18 \zeta (8) \zeta (9)+330 \zeta (6) \zeta (11)+2574 \zeta (4) \zeta (13)+12870 \zeta (2) \zeta (15)-24310 \zeta (17) \\ 10 & -110 \zeta (8) \zeta (11)-1430 \zeta (6) \zeta (13)-10010 \zeta (4) \zeta (15)-48620 \zeta (2) \zeta (17)+92378 \zeta (19)-\frac{524875 \zeta (20)}{523833} \\ \end{array}\etiqueta{2b} $$

Asymptotic behaviour

Numerically we find for large $m$ a straight exponential decay:

$$N(s_m)=1.01021 e^{-1.38663\; m}$$

Discussion

Notice that for odd $m$ the structure is simpler, at least the fraction is missing.

The coefficient series are not in OEIS.

Derivation

We consider the partial sum up to $n$ and in then take the limit $n\to\infty$.

We shall use partial summation (PS) as was shown by Jack D'Aurizio in his solution to be the method of choice.

Remember that partial summation (PS) allows to transform a sum similar to the well known partial integration (PI)

$$\int_1^n A'(x) b(x) \, dx=A(x) b(x)|_1^n -\int_1^n A(x) b'(x) \, dx$$

namely

$$\sum _{k=1}^n a_k b_k=A_n b_n- \sum _{k=1}^{n-1} A_k (b_{k+1}-b_k)\tag{3}$$

where $A_k=\sum _{j=1}^k a_j$ is the "integral" of $a_k$, and $b_{k+1}-b_k$ is the "derivative" of $b_k$.

For a given summand the choice of the factors $a_k$ and $b_k$ is to a certain degreee arbitrary and one choice can be more favorable than the other. In any case the choice determines the rest of the calculations and hence should be mentioned in the beginning.

Here we take

$$a_k = b_k = \zeta (m)-H_k^{(m)}$$

Then we find

For the "derivative"

$$b_{k+1}-b_k=H_k^{(m)}-H_{k+1}^{(m)}=-\frac{1}{(1+k)^m}\tag{4a}$$

and for the "integral"

$$A_k=\sum _{j=1}^k a(j)=H_{k+1}^{(m-1)}-(k+1) H_{k+1}^{(m)}+k \zeta (m)\tag{4b}$$

Here we have used the summation formula

$$\sum _{k=1}^n H_k^{(m)}=(n+1) H_{n+1}^{(m)}-H_{n+1}^{(m-1)}\tag{5}$$

Plugging (4) into (3) the r.h.s. of (3) is given by the sum of four terms which are written down here already in the limiting form for $n \to \infty$

$$R_1=A_n b_n \to 0\tag{6a}$$ $$R_2 = \sum _{k=1}^{\infty} \frac{k \zeta (m)}{(k+1)^m} \to \zeta (m) (\zeta (m-1)-\zeta (m))\tag{6b}$$ $$R_3 = -\sum _{k=1}^{\infty} \frac{H_k^{(m)}}{k^{m-1}}= - S_{m,m-1}\tag{6c}$$ $$R_4 = \sum _{k=1}^{\infty} \frac{H_k^{(m-1)}}{k^m}= S_{m-1,m}\tag{6d}$$

Here we have used the notation of [1] for $R_3$ and $R_4$.

These authors provide a formula of Borwein for $S_{p,q}$ which is valid for the case of odd "weight" $w=p+q$.

We are lucky here, as we have always odd weight (p+q=2m-1).

Putting the parts together we have finally

$$s_{m,2}(1) = \zeta (m) (\zeta (m-1)-\zeta (m)) - S_{m,m-1} + S_{m-1,m}\tag{7}$$

Some results have been shown above. They were checked numerically to be valid.

In order to have everything in one place here is the function $S$

$$S_{p,q}=(-1)^p \sum _{k=1}^{\left\lfloor \frac{q}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{p-1} \zeta (-2 k+p+q)+(-1)^p \sum _{k=1}^{\left\lfloor \frac{p}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{q-1} \zeta (-2 k+p+q)+\left(-\frac{1}{2} (-1)^p \binom{p+q-1}{p}-\frac{1}{2} (-1)^p \binom{p+q-1}{q}+\frac{1}{2}\right) \zeta (p+q)+\frac{1}{2} \left(1-(-1)^p\right) \zeta (p) \zeta (q)\text{/;}\text{OddQ}[p+q]$$

For the convenience of some users, here is also the Mathematica code for copying (not for reading)

S[p_, q_] :=
 (* = Sum[HarmonicNumber[k,p]/k^q,{k,1,\[Infinity]}], if p+q odd *) (Zeta[p + q] (1/2 - (-1)^p/2 Binomial[p + q - 1, p] - (-1)^p/2 Binomial[p + q - 1, q]) + (1 - (-1)^p)/2 Zeta[p] Zeta[q] + (-1)^p Sum[Binomial[p + q - 2 k - 1, q - 1] Zeta[2 k] Zeta[p + q - 2 k], {k, 1, Floor[p/2]}] + (-1)^p Sum[Binomial[p + q - 2 k - 1, p - 1] Zeta[2 k] Zeta[p + q - 2 k], {k, 1, Floor[q/2]}]) /; OddQ[p + q]

The alternating sum

I have calculated the alternating sum

$$s_a = s_{m,2}(-1) = \sum_{k\ge1} (-1)^k (\zeta(m) - H_k^{(m)})^2$$

using partial summation with the distribution $a_k = (-1)^k, b_k = (\zeta(m)-H_k^{(m)})^2$ with the (preliminary?) result

$$s_{m,2}(-1)=S^{+-}_{m,m}-\left(-2^{1-m} \zeta (m)^2-2^{-(2 m)} \zeta (2 m)+\frac{1}{2} \left(\zeta (m)^2+\zeta (2 m)\right)\right)\tag{8}$$

Where (using the definition of [1])

$$S^{+-}_{p,q} = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(p)}}{k^q}$$

I have checked numerically that (8) is correct (to a high degree).

The sum $S^{+-}_{p,q}$ has been studied extensively in [1], but alas, closed forms are given (and seem to exist) only for odd weight $w = p + p$, and we have even weigth $2m$.

Also in [2] our case, designated $\alpha_h(m,n)$ there, is circumnavigated.

For $m=1$ the closed form result is known [3]:

$$S^{+-}_{1,1} = \frac{\log ^2(2)}{2}-\frac{\zeta(2)}{2}$$

For $m=2$ the question was asked in [4] and the answer was given by Przemo. Quoting also his notation it is

$$S^{+-}_{2,2}={\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$$

It is not clear (to me) if his results really lead to closed form expressions for $m \ge 3$.

Conclusion

The question of a closed form was answered here affirmative for the non-alternating sum. For the alternating sum it could be traced back here to the reducibiity of the basic sum $S^{+-}_{m,m}$.

It remains to be seen if the confirmed reducibility for $m=1$ and $m=2$ can be extended to higher $m$.

References

[1] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf, Euler Sums and contour integral representations, Philippe Flajolet and Bruno Salvy, 1995

[2] http://www.davidhbailey.com/dhbpapers/eulsum-em.pdf, Experimental Evaluation of Euler Sums, David H. Bailey, Jonathan M. Borwein and Roland Girgensohn, 1994

[3] $S^{+-}_{1,1} = \frac{1}{12} \left(6 \log ^2(2)-\pi ^2\right)$, Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$, asked by Spivey, answered by many

[4] $S^{+-}_{2,2}$, el Cálculo de la alternancia de Euler sumas de potencias impares, preguntado por Zaid Alyafeai, las respuestas fueron dadas por Przemo

Answer 3

Voy a añadir una de respuestas separada para mostrar una forma cerrada para $s_{2,3}(1)$. Por summmation por partes

$$ \sum_{k\geq 1}\left(\zeta(2)-H_m^{(2)}\right)^3=S_1+S_2+S_3 $$ donde $$ S_1=\sum_{k\geq 1}\left[-\frac{1}{k^6}+\frac{1}{k^5}-\frac{\pi ^2}{2 k^4}+\frac{\pi ^2}{2 k^3}-\frac{\pi ^4}{12 k^2}+\frac{3 H_k^{(2)}}{k^4}-\frac{3 H_k^{(2)}}{k^3}+\frac{\pi ^2 H_k^{(2)}}{k^2}\right] $$ $$ S_2=\sum_{k\geq 1}\left[\frac{\pi ^4}{12 k}-\frac{\pi ^2 H_{k}^{(2)}}{k}+\frac{3\left(H_k^{(2)}\right)^2}{k}\right]=3\sum_{k\geq 1}\frac{\left(\zeta(2)-H_k^{(2)}\right)^2}{k}$$ $$ S_3 = -3\sum_{k\geq 1}\frac{\left(H_k^{(2)}\right)^2}{k^2}. $$ $S_1$ puede ser calculada a través de la norma de Euler sumas y $S_3$ puede ser calculada a través de los anteriores y la suma por partes: $$ S_1 = -2\zeta(6)-6\zeta(2)\zeta(3)+3\zeta(3)^2+\frac{29}{2}\zeta(5) $$ $$ S_3 = 2\zeta(6)-3\zeta(3)^2-\zeta(2)^3 $$ Acerca de $S_2$, suma por partes y Euler sumas dar $$ S_2 = 21\zeta(2)\zeta(3)-21\zeta(5)-6\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2} $$ y por Zaid el resultado descrito aquí tenemos $$\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2}=\zeta(2)\,\zeta(3)+\zeta(5). $$

De ello se deduce que también se $s_{2,3}$ tiene una forma cerrada en términos de los valores de la $\zeta$ función de: $$\boxed{ \sum_{k\geq 1}\left(\zeta(2)-H_{k}^{(2)}\right)^3 = \color{red}{9\,\zeta(2)\,\zeta(3)-\zeta(2)^3-\frac{25}{2}\zeta(5)}.} $$