21
$\begingroup$

Here is a pretty series

$$ \displaystyle \sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)=\frac{1}{2} \left(1-\ln (2\pi)+\gamma\right) \tag{*} $$

where $H_{n}:=\sum_{1}^{n} \frac{1}{k}$ are the harmonic numbers and $\gamma := \lim\limits_{n \to \infty} (H_n- \ln n)$ is the Euler constant.

$$ $$

Now just introduce a parameter in the general term of the series and you get a link with... the Riemann $\zeta$ function on the critical line!

Q 1. What proof would you give for (*)?

Q 2. What elements would you give to get the link with $\zeta\left(\frac{1}{2}+it\right)$?

$\endgroup$
  • 1
    $\begingroup$ I'm unsure what exactly you mean by "introduce a parameter in the term of the series". My first guess is that you intended that we create a power series in the parameter using the series terms as coefficients: $f(z)=\sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)z^n$. Is that correct? $\endgroup$ – David H Jul 21 '14 at 10:53
  • $\begingroup$ @David H No, it isn't. We don't create a power series. We "generalize" the term of the initial series with a parameter ... Hoping to be clear. David: I don't mean it is easy! $\endgroup$ – Olivier Oloa Jul 21 '14 at 12:07
  • $\begingroup$ A similar series is given by wolframalpha.com/input/… $\endgroup$ – Jaume Oliver Lafont Jan 11 '16 at 6:13
17
$\begingroup$

Proof of (*)

Adding the four finite sums,

$$\sum_{k=1}^{n}H_{k}=(n+1)H_{n}-n,$$

$$\sum_{k=1}^{n}\ln{k}=\ln{n!},$$

$$\sum_{k=1}^{n}\gamma=\gamma\,n,$$

$$\sum_{k=1}^{n}\frac{1}{2k}=\frac12H_{n},$$

gives us a representation of the $n$-th partial sum for the infinite series. Writing the infinite series as the limit of partial sums, we get:

$$\begin{align} S &=\sum_{k=1}^{\infty}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\ &=\lim_{n\to\infty}\sum_{k=1}^{n}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\ &=\lim_{n\to\infty}\left((n+1)H_{n}-n-\ln{n!}-\gamma\,n-\frac12H_{n}\right)\\ &=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right). \end{align}$$

Use Stirling's approximation for the factorial to obtain an asymptotic formula for the log-factorial term in the series:

$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\ \implies \ln{n!}\sim\ln{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)}=\left(n+\frac12\right)\ln{n}-n+\frac12\ln{(2\pi)}.$$

Then,

$$\begin{align} S &=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right)\\ &=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\left(n+\frac12\right)\ln{n}+n-\frac12\ln{(2\pi)}\right)\\ &=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-\gamma\,n-\left(n+\frac12\right)\ln{n}\right)-\frac12\ln{(2\pi)}\\ &=\lim_{n\to\infty}\left(n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\left(H_{n}-\ln{n}\right)\right)-\frac12\ln{(2\pi)}\\ &=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\lim_{n\to\infty}\left(H_{n}-\ln{n}\right)-\frac12\ln{(2\pi)}\\ &=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\gamma-\frac12\ln{(2\pi)}\\ &=\frac12+\frac12\gamma-\frac12\ln{(2\pi)}.~~~\blacksquare \end{align}$$


Appendix:

Using the asymptotic series for the digamma function given by Eq.16 on this Wolfram Mathworld page,

$$\begin{align} \lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right) &=\lim_{n\to\infty}n\left(\Psi{(n+1)}-\ln{n}\right)\\ &=\lim_{n\to\infty}n\left(\frac{1}{2n}-\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell}}\right)\\ &=\frac12-\lim_{n\to\infty}\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell-1}}\\ &=\frac12. \end{align}$$

$\endgroup$
  • $\begingroup$ Thanks for this correct answer to the first part of the question. (+1) $\endgroup$ – Olivier Oloa Jul 21 '14 at 10:40
  • $\begingroup$ Do you mean $\log k $ in the second sum? $\endgroup$ – YoTengoUnLCD Jan 26 '16 at 4:03
  • $\begingroup$ @YoTengoUnLCD Yes, you are right. The formula has now been corrected. :) $\endgroup$ – David H Jan 28 '16 at 4:07
9
$\begingroup$

Observe that $$ H_{n}-\ln n-\gamma -\frac{1}{2n} = \psi (n) - \ln n + \frac{1}{2n} $$ where $\psi := \Gamma'/\Gamma$ is the digamma function, using $\displaystyle \psi (n)= H_{n-1}-\gamma = H_n-\gamma- \frac{1}{n}$, $n\geq 1$.

Our initial series thus rewrites $$ \sum_{n=1}^{\infty} \left( \psi(n )- \log n + \frac{1}{2n}\right) = \frac{\gamma}{2} - \frac{1}{2}\ln(2\pi)+ \frac{1}{2}, $$ (proved by David H).

Then consider the one parameter series $$ \sum_{n=1}^{\infty}\left(\psi(n \alpha)- \log (n \alpha) + \frac{1}{2n \alpha}\right), \quad \alpha >0. $$ We have the following result.

Theorem 1. Let $\alpha$ and $\beta$ be positive real numbers such that $ \alpha\beta=1$.

Then \begin{align} &\sqrt{\alpha}\left\{\frac{\gamma-\log(2\pi\alpha)}{2\alpha}+ \sum_{n=1}^{\infty}\left(\psi(n \alpha)- \log (n \alpha) + \frac{1}{2n \alpha}\right)\right\}\\ = & \sqrt{\beta}\left\{\frac{\gamma-\log(2\pi\beta)}{2\beta}+\sum_{n=1}^{\infty}\left(\psi(n \beta)- \log (n \beta) + \frac{1}{2n \beta}\right)\right\} \\ = &-\frac{1}{\pi^{3/2}}\int_0^{\infty}\left|\xi\left(\frac{1}{2}+\frac{it}{2}\right)\Gamma\left(\frac{-1+it}{4}\right)\right|^2 \frac{\cos\left(\frac{t}{2}\log\alpha\right)}{1+t^2}dt, \end{align}

where $$ \xi(s):=\frac{s(s-1)}{2} \displaystyle \pi^{-s/2}\:\Gamma(\frac{s}{2})\zeta(s)$$ and where $\zeta$ is the Riemann zeta function.

Now express $\displaystyle \left|\xi\left(\frac{1}{2}+\frac{it}{2}\right)\right|^2 $ in terms of $\left|\zeta \left(\frac{1}{2}+ it\right)\right|^2$ and you obtain the evocated link.

Theorem 1 is due to Ramanujan and one may find a recent proof here.

Here is a related result I have found.

Theorem 2. Let $\Re \alpha >0$.

Then $$ \sum_{n=1}^{\infty} \! \left(\! \psi(\alpha n )- \log (\alpha n ) + \frac{1}{2 \alpha n }\! \right)\! =\! \displaystyle \frac{1+\gamma-\log(2\pi)}{2} \\ -\int_{0}^{1}\left(\frac{1}{\alpha (1-x^{1/\alpha})}-\frac{1}{1-x}+\frac{1}{2}-\frac{1}{2\alpha}\!\right)\!\frac{\mathrm{d}x}{1-x}.$$

Thanks.

$\endgroup$
  • $\begingroup$ It looks like for $\alpha\in\mathbb Q$ the last integral has a form $p+q\ln\alpha+\pi\tau$, where $p,q\in\mathbb Q$, and $\tau$ is an algebraic number. I could not find general expressions for those coefficients. $\endgroup$ – Vladimir Reshetnikov Dec 24 '14 at 19:38
7
$\begingroup$

Since $$ 1+\sum_{k=2}^n\left(\frac1k-\log\left(\frac{k}{k-1}\right)\right)=H_n-\log(n) $$ we have $$ \begin{align} H_n-\log(n)-\gamma &=\sum_{k=n+1}^\infty\left(\log\left(\frac{k}{k-1}\right)-\frac1k\right)\\ &=\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac1{k+1}\right) \end{align} $$ Furthermore, $$ \frac1{2n}=\sum_{k=n}^\infty\frac12\left(\frac1k-\frac1{k+1}\right) $$ Therefore, $$ H_n-\log(n)-\gamma-\frac1{2n}=\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right) $$ Summing, we have $$ \begin{align} &\sum_{n=1}^\infty\left(H_n-\log(n)-\gamma-\frac1{2n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\ &=\sum_{k=1}^\infty\sum_{n=1}^k\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\ &=\sum_{k=1}^\infty k\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right) \end{align} $$ and $$ \begin{align} &\sum_{k=1}^nk\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\ &=\sum_{k=2}^{n+1}(k-1)\log(k)-\sum_{k=1}^nk\log(k)-\sum_{k=1}^n\left(1-\frac1{2(k+1)}\right)\\ &=\color{#C00000}{n\log(n+1)}\color{#00A000}{-\log(n!)-n}\color{#0000F0}{+\frac12(H_{n+1}-1)}\\ &=\color{#C00000}{n\log(n)+1+O\!\left(\frac1n\right)}\color{#00A000}{-\left(n+\frac12\right)\log(n)-\frac12\log(2\pi)+O\!\left(\frac1n\right)}\\ &\color{#0000F0}{+\frac12\log(n)+\frac12\gamma-\frac12+O\!\left(\frac1n\right)}\\ &=\frac12\left(1+\gamma-\log(2\pi)\right)+O\!\left(\frac1n\right) \end{align} $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\left(H_n-\log(n)-\gamma-\frac1{2n}\right)=\frac{1+\gamma-\log(2\pi)}2} $$

$\endgroup$
  • 3
    $\begingroup$ Nice proof! (+1) $\endgroup$ – Olivier Oloa Jan 28 '16 at 16:00
0
$\begingroup$

This is not an answer but contains information that may be useful in building one.

There are closed form expressions for $H_n-log\left(n\right)$ and $\gamma$ that come from generalized Mercator series.

$$ H_n-log\left(n\right)=1-\sum_{k=1}^{\infty}\left(\sum_{i=nk+1}^{n(k+1)}\frac{1}{i}-\frac{1}{k+1}\right) $$ https://math.stackexchange.com/a/1602945/134791

$$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right) $$

https://math.stackexchange.com/a/1591256/134791

$\endgroup$
  • $\begingroup$ Thank you for reviewing and commenting, @kamil09875. I added the intended warning "this is not an answer" as soon as I noticed downvotes. However, I believe this adds relevant information and is too long for a comment. Having seen other "non-answers" written in the answer box rather than as a comment for length reasons, it seemed appropriate to me in this case, as well. Would you move this to a comment or simply delete it? $\endgroup$ – Jaume Oliver Lafont Jan 26 '16 at 9:14
  • 1
    $\begingroup$ If a comment is too long, attach it as an answer but make sure that you include a proper message. It will prevent it from reviewing as "not an answer". I've compensated the score of this post. $\endgroup$ – Kamil Jarosz Jan 26 '16 at 10:37

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.