Mostrando que $\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}$

Question

Considere esta doble suma$(1)$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}\tag1$ $ Donde$H_n$ es el n-ésimo armónico

Un intento:

Reescribir$(1)$ as

ps

Recordar$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\left(H_1+{H_2\over 2}+{H_3\over 3}+\cdots+{H_{k+1}\over k+1}\right)\tag2$ $

No está seguro de cómo continuar

Answer 1

Podemos observar que la $$ \sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1} H_n x^n = -\frac{\log(1-x)}{1-x}\tag{1} $$ por lo tanto $$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n}=\frac{\log^2(1-x)}{2x},\qquad \sum_{n\geq 1}\frac{H_{n+1}}{n+1} x^{n}=\frac{\log^2(1-x)+2\text{Li}_2(x)}{2x}-1 \tag{2}$$ y podemos considerar que lo que el operador $$ T_n=\sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{[x^k]}{k+1} \tag{3} $$ hace a una analítica de la función en una vecindad de cero. Esto es estrictamente relacionados con el binomio de transformación (y con los números de Stirling de primera especie , dando la serie de Taylor de $\log(1-x)^k$), por lo tanto, para resolver la pregunta es suficiente para calcular de una forma cerrada para $$ \sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{1}{(k+1)^3} = \frac{1}{2}\int_{0}^{1}\sum_{k=0}^{n}(-1)^k \binom{n}{k} x^k \log^2(x)\,dx = \frac{1}{2}\int_{0}^{1}(1-x)^n \log^2(x)\,dx$$ donde la última integral es $$ \frac{d^2}{d\alpha^2}\left.\int_{0}^{1}(1-x)^n x^{\alpha}\,dx\,\right|_{\alpha=0^+}=\frac{H_{n+1}^2+H_n^{(2)}}{2n+2}.\tag{4}$$ Para finalizar la prueba, es suficiente para aplicar la sumación por partes $$ \sum_{j=0}^{k}\frac{H_{j+1}}{j+1} = H_{k+1}^{2}-\sum_{j=0}^{k-1}\frac{H_{j+1}}{j+1}.\tag{5} $$ Larga historia corta: OP identidad a partir de la aplicación de la binomial transformar a la serie la definición de $\zeta(3)$ y explotar mis $(4)$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 0}^{n}\pars{-1}^{k}{n\elegir k}{1 \over k + 1} \sum_{j = 0}^{k}{H_{j + 1} \over j + 1} = {1 \over \pars{n + 1}^{3}}:\ {\grande ?}}$.

Con el Número Armónico de Generación de Función $\ds{\,\mc{H}\pars{z} \equiv -\,{\ln\pars{1 - z} \over1 - z} = \sum_{n = 0}^{\infty}H_{n}z^{n}}$, vamos a considerar la sobre-$\ds{j}$ suma: \begin{align} \sum_{j = 0}^{k}{H_{j + 1} \over j + 1} & = \sum_{j = 0}^{k}{\bracks{z^{\,j + 1}}\mc{H}\pars{z} \over j + 1} = \bracks{z^{0}}\mc{H}\pars{z}\sum_{j = 0}^{k}{\pars{1/z}^{\,j + 1} \over j + 1} = \bracks{z^{0}}\mc{H}\pars{z}\sum_{j = 0}^{k}\int_{0}^{1/z}x^{\,j}\,\dd x \\[5mm] & = \bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z} {x^{k + 1} - 1 \over x - 1}\,\dd x \end{align}

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A continuación, \begin{align} &\sum_{k = 0}^{n}\pars{-1}^{k}{n\choose k}{1 \over k + 1} \sum_{j = 0}^{k}{H_{j + 1} \over j + 1} = -\sum_{k = 0}^{n}{n\choose k}{\pars{-1}^{k + 1} \over k + 1} \braces{\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z} {x^{k + 1} - 1 \over x - 1}\,\dd x} \\[5mm] = &\ \bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z} \bracks{% \sum_{k = 0}^{n}{n\choose k}{\pars{-x}^{k + 1} \over k + 1} - \sum_{k = 0}^{n}{n\choose k}{\pars{-1}^{k + 1} \over k + 1}}{\dd x \over 1 - x} \\[5mm] = &\ \bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z} \bracks{\pars{1 - x}^{n + 1} \over n + 1}{\dd x \over 1 - x} = {1 \over n + 1}\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}\pars{1 - x}^{n}\,\dd x \\[5mm] = &\ {1 \over n + 1}\bracks{z^{0}}\mc{H}\pars{z} {1 + \pars{-1}^n\pars{1/z - 1}^{n + 1} \over n + 1} \\[5mm] = &\ {1 \over \pars{n + 1}^{2}}\braces{\vphantom{\Large A}% \bracks{z^{0}}\mc{H}\pars{z} + \pars{-1}^{n}\bracks{z^{n + 1}}\mc{H}\pars{z}\pars{1 - z}^{n + 1}} \\[5mm] = &\ {\pars{-1}^{n + 1} \over \pars{n + 1}^{2}} \bracks{z^{n + 1}}\pars{1 - z}^{n}\ln\pars{1 - z} = {\pars{-1}^{n + 1} \over \pars{n + 1}^{2}}\bracks{z^{n + 1}} \left.\partiald{\pars{1 - z}^{n + \mu}}{\mu}\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ {\pars{-1}^{n + 1} \over \pars{n + 1}^{2}} \left.\partiald{}{\mu}{n + \mu \choose n + 1}\pars{-1}^{n + 1} \right\vert_{\ \mu\ =\ 0} = \bbx{\ds{1 \over \pars{n + 1}^{3}}} \end{align}

Tenga en cuenta que

\begin{align} \partiald{}{\mu}{n + \mu \choose n + 1} & = {n + \mu \choose n + 1}\pars{H_{n + \mu} - H_{\mu - 1}} = {-\mu \choose n + 1}\pars{-1}^{n + 1}\pars{H_{n + \mu} - H_{\mu - 1}} \\[5mm] & = \pars{-1}^{n + 1}\, {\Gamma\pars{1 - \mu} \over \pars{n + 1}!\,\Gamma\pars{-\mu - n}} \bracks{H_{n + \mu} - H_{-\mu} + \pi\cot\pars{\pi\mu}} \\[5mm] & = \pars{-1}^{n + 1}\, {\Gamma\pars{1 - \mu}\bracks{H_{n + \mu} - H_{-\mu} + \pi\cot\pars{\pi\mu}} \over \pars{n + 1}!\pars{\pi/\braces{\Gamma\pars{n + 1 + \mu} \sin\pars{\pi\bracks{n + 1 + \mu}}}}} \\[5mm] & = \underbrace{\Gamma\pars{1 - \mu}}_{\ds{\to\ 1\ \mrm{as}\ \mu\ \to\ 0}}\,\ \underbrace{{\Gamma\pars{n + 1 + \mu} \over \pars{n +1}!}} _{\ds{\to\ {1 \over n + 1}\ \mrm{as}\ \mu\ \to\ 0}}\,\ \underbrace{\bracks{{\sin\pars{\pi\mu}\pars{H_{n + \mu} - H_{-\mu}} \over \pi} + \cos\pars{\pi\mu}}}_{\ds{\to\ 1\ \mrm{as}\ \mu\ \to\ 0}} \end{align}

Answer 3

Buscamos mostrar que

$$\sum_{k=0}^n (-1)^k {n\elegir k} \frac{1}{k+1} \sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}.$$

Este es

$$\sum_{k=0}^n (-1)^k {n+1\elegir k+1} \frac{k+1}{n+1} \frac{1}{k+1} \sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}$$

o

$$\sum_{k=0}^n (-1)^k {n+1\elegir k+1} \sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^2}.$$

El LHS es

$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} \sum_{k=j}^n (-1)^k {n+1\choose k+1}.$$

Escrito

$${n+1\elegir k+1} = {n+1\elegir n-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+1} \; dz$$

podemos obtener el rango de control (se desvanece para $k\gt n$) por lo que se puede escribir para la interior de la suma

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n+1} \sum_{k\ge j} (-1)^k z^k \; dz \\ = (-1)^j \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} (1+z)^{n+1} \sum_{k\ge 0} (-1)^k z^k \; dz \\ = (-1)^j \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} (1+z)^{n+1} \frac{1}{1+z} \; dz \\ = (-1)^j \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} (1+z)^{n} \; dz = (-1)^j {n\elegir n-j} = (-1)^j {n\elegir j}.$$

Por lo tanto, tiene que demostrar que

$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j {n\elegir j} = \frac{1}{(1+n)^2}$$

o, alternativamente,

$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j \frac{j+1}{n+1} {n+1\elegir j+1} = \frac{1}{(1+n)^2}$$

que es

$$\sum_{j=0}^n H_{j+1} (-1)^j {n+1\elegir j+1} = \frac{1}{1+n}$$

El LHS es

$$\sum_{j=0}^n (-1)^j {n+1\elegir j+1} \sum_{q=1}^{j+1} \frac{1}{q} = \sum_{j=0}^n (-1)^j {n+1\elegir j+1} \sum_{q=0}^{j} \frac{1}{p+1} \\ = \sum_{q=0}^n \frac{1}{p+1} \sum_{j=q}^n (-1)^j {n+1\elegir j+1}.$$

Nos re-utilizar el cálculo de antes

$$\sum_{q=0}^n \frac{1}{p+1} (-1)^p {n\elegir q} = \sum_{q=0}^n \frac{1}{p+1} (-1)^q \frac{q+1}{n+1} {n+1\elegir q+1} \\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^p {n+1\elegir q+1}.$$

Hemos reducido la demanda a

$$\sum_{q=0}^n (-1)^q {n+1\choose q+1} = 1$$

que tiene por la inspección o por escrito

$$- \sum_{q=1}^{n+1} (-1)^p {n+1\elegir q} = 1 - \sum_{q=0}^{n+1} (-1)^p {n+1\elegir q} = 1 - (1-1)^{n+1} = 1.$$