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I know how to prove

$$\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$$ by tackling it with the beta function.

I was actually wondering if there is a proof of this fact without using the property of the Beta function $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

pisco
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Interesting problems
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2 Answers2

11

We have \begin{align} f_n&=\color{blue}{\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{1}{k}}\\ &=\sum_{k=1}^{n}(-1)^{k-1}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)\frac{1}{k}\\ &=f_{n-1}+\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}\frac{1}{k}\\ &=f_{n-1}-\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}\\ &=f_{n-1}-\frac{1}{n}\left((1-1)^n-1\right)\\ &=f_{n-1}+\frac{1}{n}\\ &\,\,\color{blue}{=H_n} \end{align}

Note: This can be found for instance as Example 3, section 1.2 in Combinatorial Identities by John Riordan.

Another approach is based upon generating functions and the Euler transform of a generating series $A(z)$ which is given as \begin{align*} A(z)=\sum_{n= 0}^\infty a_nz^n&\quad\longrightarrow\quad \frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_k\right)z^n\tag{1}\\ a_n\quad&\quad\longrightarrow\quad\qquad \sum_{k=0}^n\binom{n}{k}a_k& \end{align*}

Applying the Euler transform (1) to the left-hand side of the identity \begin{align*} \sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n \end{align*} we set $a_k=\frac{(-1)^{k-1}}{k}$ and obtain as generating function \begin{align*} A(z)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}z^k=\ln(1+z)\tag{2} \end{align*}

The Euler transform of (2) is \begin{align*} \frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\sum_{n=1}^\infty\left(\color{blue}{\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k}}\right)z^n \end{align*} On the other hand we have \begin{align*} \frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\frac{1}{1-z}\ln\left(\frac{1}{1-z}\right)\\ &=-\frac{1}{1-z}\ln(1-z)\\ &=\left(\sum_{k=0}^\infty z^k\right)\left( \sum_{l=1}^\infty \frac{1}{l}z^l\right)\\ &=\sum_{n=1}^\infty\left(\sum_{{k+l=1}\atop{k\geq 0,l\geq 1}}^n \frac{1}{l}\right)z^n\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{1}{k}z^n\\ &=\sum_{n=1}^\infty\color{blue}{H_n}z^n\\ \end{align*} and the claim follows.

Note: A proof of the Euler transformation formula can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev.

Markus Scheuer
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  • can you please explain me How I Get $\displaystyle f_{n} = f_{n_1}+\frac{1}{n} = H_{n}$, because when i open i am getting $\displaystyle f_{n} = f_{n-(n-1)}+\frac{1}{n-(n-2)}+\frac{1}{n-(n-3)}+\cdots \cdots +\frac{1}{n-1}+\frac{1}{n}..$ Now i am getting $f_{1} = n.$ Then how can we say $f_{n}=H_{n}.$ please explain me. Thanks – DXT Dec 11 '18 at 10:30
  • @DurgeshTiwari: I've added a step which might be helpful. – Markus Scheuer Dec 11 '18 at 11:42
  • @DurgeshTiwari: We have $f_n=f_{n-1}+\frac{1}{n}=\left(f_{n-2}+\frac{1}{n-1}\right)+\frac{1}{n}=\left(f_{n-3}+\frac{1}{n-2}\right)+\frac{1}{n-1}+\frac{1}{n}=\cdots=H_n$. – Markus Scheuer Dec 11 '18 at 11:49
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 1}^{n}{\pars{-1}^{k - 1} \over k}{n \choose k} = H_{n}:\ {\LARGE ?}.\quad H_{z}:\ Harmonic\ Number}$.

\begin{align} &\bbox[10px,#ffd]{\sum_{k = 1}^{n}{\pars{-1}^{k - 1} \over k}{n \choose k}} = \sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k}{n \choose n - k} = \sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k}\bracks{z^{n - k}} \pars{1 + z}^{n} \\[5mm] = &\ \bracks{z^{n - 1}} \pars{1 + z}^{n}\sum_{k = 1}^{\infty}{\pars{-z}^{k - 1} \over k} = \bracks{z^{n - 1}}\pars{1 + z}^{n}\sum_{k = 1}^{\infty} \pars{-z}^{k - 1}\int_{0}^{1}t^{k - 1}\,\dd t \\[5mm] = &\ \bracks{z^{n - 1}}\pars{1 + z}^{n}\int_{0}^{1} \sum_{k = 1}^{\infty} \pars{-zt}^{k - 1}\,\dd t = \bracks{z^{n - 1}}\pars{1 + z}^{n}\int_{0}^{1} {1 \over 1 - \pars{-zt}}\,\dd t \\[5mm] = &\ \bracks{z^{n - 1}}\pars{1 + z}^{n}\,{\ln\pars{1 + z} \over z} = \bracks{z^{n}}\, \lim_{\nu\ \to\ n}\partiald{\pars{1 + z}^{\nu}}{\nu} = \lim_{\nu\ \to\ n}\partiald{}{\nu}{\nu \choose n} \\[5mm] = &\ {1 \over n!}\lim_{\nu\ \to\ n}\partiald{}{\nu} \bracks{\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu - n + 1}} \\[5mm] = &\ {1 \over n!}{\Gamma\pars{n + 1}\Psi\pars{n + 1}\Gamma\pars{1} - \Gamma\pars{1}\Psi\pars{1}\Gamma\pars{n + 1} \over \Gamma^{2}\pars{1}} \\[5mm] = &\ \Psi\pars{n + 1} + \gamma = \bbx{H_{n}} \end{align}

Felix Marin
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