Calculus – Sequences and series – The harmonic numbers

Introducing the harmonic numbers\({}^{(*)}\) \(H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}\) for \(n=1,2,3,\ldots\), we obtain that

\[\sum_{n=1}^{\infty}\frac{H_n}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^n\frac{1}{k} =\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\frac{1}{kn^2}=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{1}{k(n+k)^2} =\sum_{k=1}^{\infty}\frac{1}{k^3}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k(n+k)^2}.\]

Now we have

\begin{align*} \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k(n+k)^2} &=\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\left(\frac{1}{k(n+k)^2}+\frac{1}{n(k+n)^2}\right) =\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{k+n}{kn(k+n)^2}\\[2.5mm] &=\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{kn(k+n)} =\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{1}{k}-\frac{1}{k+n}\right)\\[2.5mm] &=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+n}\right) =\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^n\frac{1}{k}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_n}{n^2}. \end{align*}

So we conclude that \(\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^2}=\sum_{k=1}^{\infty}\frac{1}{k^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_n}{n^2}\), which implies that\({}^{(**)}\) \(\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)\).

The harmonic numbers satisfy the recursion relation \(H_{n+1}=H_n+\displaystyle\frac{1}{n+1}\) for \(n=1,2,3,\ldots\). Now suppose that \(G(x)=\displaystyle\sum_{n=1}^{\infty}H_nx^n\), then we have \[G(x)=\sum_{n=0}^{\infty}H_{n+1}x^{n+1}=\sum_{n=1}^{\infty}H_nx^{n+1}+\sum_{n=0}^{\infty}\frac{1}{n+1}x^{n+1} =xG(x)-\ln(1-x)\quad\Longrightarrow\quad G(x)=-\frac{\ln(1-x)}{1-x}.\]

Since \(\displaystyle\lim\limits_{n\to\infty}\frac{H_{n+1}}{H_n}=\lim\limits_{n\to\infty}\left(1+\frac{1}{(n+1)H_n}\right)=1\) we conclude that the radius of convergence is \(1\). Another way to obtain this result is by using \(\displaystyle\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) and \(\displaystyle-\ln(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k}\) for \(|x| < 1\):

\[-\frac{\ln(1-x)}{1-x}=\sum_{n=0}^{\infty}x^n\sum_{k=1}^{\infty}\frac{x^k}{k}=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{x^{n+k}}{k} =\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{x^{n+k}}{k}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\frac{x^n}{k} =\sum_{n=1}^{\infty}\left(\sum_{k=1}^n\frac{1}{k}\right)x^n=\sum_{n=1}^{\infty}H_nx^n,\quad |x| < 1.\]

For instance, this implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{2^n}=G(\tfrac{1}{2})=-\frac{\ln\left(\frac{1}{2}\right)}{1-\frac{1}{2}}=2\ln(2)\).

Differentiation leads to

\[\sum_{n=1}^{\infty}nH_nx^{n-1}=\frac{1}{(1-x)^2}-\frac{\ln(1-x)}{(1-x)^2}=\frac{1-\ln(1-x)}{(1-x)^2},\quad|x| < 1.\]

For instance, this implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{nH_n}{2^n}=\frac{1}{2}\cdot\frac{1-\ln(\frac{1}{2})}{(\frac{1}{2})^2}=2\left(1+\ln(2)\right)\).

Differentiating once more gives

\[\sum_{n=2}^{\infty}n(n-1)H_nx^{n-2}=\frac{1-x+2(1-x)\left(1-\ln(1-x)\right)}{(1-x)^4}=\frac{1+2\left(1-\ln(1-x)\right)}{(1-x)^3} =\frac{3-2\ln(1-x)}{(1-x)^3},\quad|x| <1 .\]

For instance, this implies that \(\displaystyle\sum_{n=2}^{\infty}\frac{n(n-1)H_n}{2^n}=\frac{1}{4}\cdot\frac{3-2\ln(\frac{1}{2})}{(\frac{1}{2})^3}=6+4\ln(2)\).

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\({}^{(*)}\) So the sequence of harmonic numbers is:

\[H_1=1,\quad H_2=1+\frac{1}{2}=\frac{3}{2},\quad H_3=\frac{3}{2}+\frac{1}{3}=\frac{11}{6},\quad H_4=\frac{11}{6}+\frac{1}{4}=\frac{25}{12}, \quad H_5=\frac{25}{12}+\frac{1}{5}=\frac{137}{60},\quad\ldots.\]

\({}^{(**)}\) Here \(\zeta(s)\) denotes the Riemann zeta function.

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Last modified on March 15, 2021