# Sum of Harmonic Number error over all $n$

It is well known that

$\displaystyle \mathcal{H}_n - \log{\left(n+\frac{1}{2}\right)} - \gamma = \frac{1}{24n^2} -\frac{1}{24n^3} + \mathcal{O}\left(\frac{1}{n^4}\right)$

Thus I pose the following problem:

Prove the following equality

$\displaystyle \sum_{n=1}^\infty \left[ \mathcal{H}_n - \log{\left(n+\frac{1}{2}\right)} - \gamma \right] = \gamma + \frac{1-3\log{2}}{2}$

Related remarks:

Are there any similar approximations for the generalised harmonic numbers that yield errors of $\mathcal{O}(n^{-2})$ or smaller?

If so, do the sums have closed forms like this one?

• Do you know the equality to be true? If so, how? – Gerry Myerson Dec 8 '16 at 6:04
• Mathematica verifies it. – Jack Tiger Lam Dec 8 '16 at 6:04
• @GerryMyerson Yes, it is true. See my solution for the details. ;-)) – Mark Viola Dec 8 '16 at 6:57

## 1 Answer

From the expansion expressed in the OP, we have

$$NH_N-N\log\left(N+\frac12\right)-N\gamma =O\left(\frac1N\right) \tag1$$

We will exploit the expansion in $(1)$ in the following development.

We wish to evaluate the series $S$ as given by

$$S=\sum_{n=1}^\infty \left(H_n-\log\left(n+\frac12\right)-\gamma\right)$$

To proceed, we examine the terms in the sequence of partial sums $S_N$

$$S_N=\sum_{n=1}^N \left(H_n-\log\left(n+\frac12\right)-\gamma\right) \tag 2$$

The first term on the right-hand side of $(2)$ can be written as

\begin{align} \sum_{n=1}^N H_n&=\sum_{n=1}^N \sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^N \left(\frac{1}{k} \sum_{n=k}^{N}(1)\right)\\\\ &=\sum_{n=1}^N \frac{N-n+1}{n}\\\\ &=(N+1)H_N-N \tag 3 \end{align}

The second term on the right-hand side of $(2)$ can be written as

\begin{align} \sum_{n=1}^N \log\left(n+\frac12\right)&=\log\left((2N+1)!!\right)-N\log(2)\\\\ &=\log\left(\frac{(2N+1)!}{2^N\,N!}\right)-N\log(2)\\\\ &=\log\left(\frac{\sqrt{2\pi(2N+1)}\left(\frac{2N+1}{e}\right)^{2N+1}}{\sqrt{2\pi N}\left(\frac{N}{e}\right)^N}\right)-2N\log(2)+O\left(\frac1N\right)\\\\ &=\frac12 \log\left(2+\frac1N\right)\\\\ &+(2N+1)\log\left(2N+1\right)-N\log(N)\\\\ &-(N+1)-2N\log(2)+O\left(\frac1N\right)\\\\ &=\frac12 \log(2)\\\\ &+(2N+1)\log(2)+(N+1)\log(N)+1\\\\ &-(N+1)-2N\log(2)+O\left(\frac1N\right)\\\\ &=\frac32\log(2)+(N+1)\log(N)-N+O\left(\frac1N\right)\tag 4 \end{align}

Using $(3)$ and $(4)$ in $(2)$ reveals

\begin{align} S&=\lim_{N\to \infty}S_N\\\\ &=\color{blue}{\lim_{N\to\infty}N\left(H_N-\log\left(N+\frac12\right)-\gamma\right)}\\\\ &+\color{red}{\lim_{N\to\infty}(H_N-\log(N))}\\\\ &+\color{green}{\lim_{N\to\infty}N\left(\log\left(N+\frac12\right)-\log(N)\right)}\\\\ &-\frac32\log(2)+\lim_{N\to\infty}\left(O\left(\frac1N\right)\right)\\\\ &=\color{blue}{0}+\color{red}{\gamma}+\color{green}{\frac12}+0\\\\ &=\gamma +\frac{1-3\log(2)}{2} \end{align}

as was to be shown!