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?

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

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!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.