I am looking for a derivation of the following sum: $$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$ My current derivation(s) uses the zeta function at negative integers (and or Stirling approximation/ the derivative of $\zeta'(0)$). I want to avoid those.

How I got an answer was via regularization of $$-\sum_{i=1}^{\infty}\frac{\zeta(-i)}{i}=\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+ \frac{1}{n}\right)-1\bigg)$$ My own other try was rewriting it via: $$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=\sum_{k=2}^{\infty} \zeta(k)(-1)^k \bigg(\frac{1}{k+1}-\frac{1}{2k}\bigg)$$ If this works I am already happy. If there's another simple way I'd love to hear it as well.

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    $\begingroup$ This doesn't make any sense because $n$ is an index variable $\endgroup$ May 10 '20 at 12:14
  • $\begingroup$ Should it be a summation on RHS? $\endgroup$
    – Henry Lee
    May 10 '20 at 12:19
  • $\begingroup$ I made a typo the rhs should be not n but pi. I am sorry. $\endgroup$
    – Gerben
    May 10 '20 at 13:08

One has \begin{align*} & \sum\limits_{n = 1}^\infty {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - t}}{{n + t}}dt} } = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{n + t}}dt} } \\ & = \sum\limits_{n = 1}^\infty {\int_{n - 1}^n {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} } = \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} . \end{align*} Now, by the Euler--Maclaurin formula, $$ \log k! = \left( {k + \frac{1}{2}} \right)\log k- k + C + \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + k}}dt} $$ with some constant $C$. It can be shown that the integral is $\mathcal{O}(k^{-1})$ and so by Stirling's formula (or the Wallis product), $C=\frac{1}{2}\log (2\pi )$. Thus \begin{align*}\sum\limits_{n = 1}^\infty {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} & = \log 1! - \left( {\left( {1 + \frac{1}{2}} \right)\log 1 - 1 + \frac{1}{2}\log (2\pi )} \right) \\ &= 1 - \frac{1}{2}\log (2\pi ). \end{align*}

  • $\begingroup$ I got here, trying to derive Stirlings formula. That's why I wanted to avoid it. Hoping there was an easier way to solve this sum. Why wallis product? $\endgroup$
    – Gerben
    May 10 '20 at 13:44
  • $\begingroup$ The use of Wallis product is a common way to determine the constant $\sqrt{2\pi}$ in Stirling's formula. The series you are looking at is a special case of the Gudermann series. See Chapter 2, §4 in "Classical Topics in Complex Function Theory" by Reinhold Remmert. $\endgroup$
    – Gary
    May 10 '20 at 15:05
  • $\begingroup$ Thanks. It's really not that exciting indeed. Think i actually seen it before. I really like your integral technique though $\endgroup$
    – Gerben
    May 10 '20 at 21:45

New Answer. Let $S_N$ denote the partial sum for the first $N$ terms. Then $S_N$ is related to the Stirling's Formula by the following computation:

\begin{align*} S_N &= \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log(n+1) - \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log n - N \\ &= \left(N+\frac{1}{2}\right)\log (N+1) - \log (N!) - N. \end{align*}

Now we consider $e^{-S_N}$ instead. Using the formula $\int_{0}^{\infty}x^{n}e^{-sx}\,\mathrm{d}x=\frac{n!}{s^{n+1}}$,

\begin{align*} \exp(-S_N) &= \frac{N!e^{N}}{(N+1)^{N+\frac{1}{2}}} \\ &= \frac{N^{N+1}}{(N+1)^{N+\frac{1}{2}}} \int_{0}^{\infty} x^N e^{-N(x-1)} \, \mathrm{d}x \\ &= \frac{1}{(1+\frac{1}{N})^{N+\frac{1}{2}}} \int_{-\infty}^{\infty} \left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u} \, \mathrm{d}u, \end{align*}

where we utilized the substitution $x=1+\frac{u}{\sqrt{N}}$ in the last step and $x_{+}:=\max\{0,x\}$ denotes the positive part of $x$. Then, taking limit as $N\to\infty$ and assuming for a moment that the order of limit and integral can be swapped, we get

\begin{align*} \lim_{N\to\infty} \exp(-S_N) &= \biggl( \lim_{N\to\infty} \frac{1}{(1+\frac{1}{N})^{N+\frac{1}{2}}} \biggr) \int_{-\infty}^{\infty} \lim_{N\to\infty} \left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u} \, \mathrm{d}u \\ &= \frac{1}{e} \int_{-\infty}^{\infty} e^{-u^2/2} \, \mathrm{d}u = \frac{\sqrt{2\pi}}{e}. \end{align*}

Here, the last step follows from the gaussian integral. Therefore

$$ \sum_{n=1}^{\infty} \left[ \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1 \right] = \lim_{N\to\infty} S_N = 1 - \log\sqrt{2\pi} $$

provided the interchange of limit and integral is justified. For this, we note the following inequality:

$$ \log(1+x) \leq x - \frac{x^2}{2(1+x_+)}, \qquad x > -1 $$

From this, we deduce that

$$ \left(1 + \frac{u}{\sqrt{N}}\right)_{+}^N e^{-\sqrt{N}u} \leq \exp\left(-\frac{u^2}{2(1+u_+)}\right) $$

holds for all $N\geq 1$ and for all $u \in \mathbb{R}$. Therefore the dominated convergence theorem is applicable and the desired step is justified, completing the proof.

Old Answer. The sum converges absolutely by the Limit Comparison Test with $\zeta(2)$. Now for each given $n \geq 1$,

\begin{align*} \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1 &= \left(n+\frac{1}{2}\right)\left(\sum_{j=1}^{\infty}\frac{(-1)^{j-1}}{jn^j} \right)-1\\ &= - \frac{1}{4n^2} + \left(n+\frac{1}{2}\right)\sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{jn^j}\\ &= - \frac{1}{4n^2} + \sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{j}\left(\frac{1}{n^{j-1}}+\frac{1}{2n^j}\right). \end{align*}

Using the formula $\int_{0}^{\infty}x^{s-1}e^{-nx}\,\mathrm{d}x=\frac{\Gamma(s)}{n^s}$, this may be recast as

\begin{align*} &= \int_{0}^{\infty}\left[ - \frac{x}{4} + \sum_{j=3}^{\infty}\frac{(-1)^{j-1}}{j}\left( \frac{x^{j-2}}{(j-2)!} + \frac{x^{j-1}}{2(j-1)!} \right)\right] e^{-nx}\, \mathrm{d}x \\ &= \int_{0}^{\infty} \left( \frac{1}{x} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)(1-e^{-x}) \right) e^{-nx} \, \mathrm{d}x. \end{align*}

Summing this for $n = 1, 2, \dots$, we get

\begin{align*} S &:= \sum_{n=1}^{\infty} \left[ \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1 \right] \\ &= \int_{0}^{\infty} \left( \frac{1}{x} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)(1-e^{-x}) \right) \frac{1}{e^x - 1} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \left( \frac{1}{x(e^x - 1)} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)e^{-x} \right) \, \mathrm{d}x. \end{align*}

To compute the right-hand side, we consider the following regularization:

\begin{align*} S(s) &:= \int_{0}^{\infty} \left( \frac{1}{x(e^x - 1)} - \left(\frac{1}{2x}+\frac{1}{x^2}\right)e^{-x} \right) x^s \, \mathrm{d}x \\ &= \int_{0}^{\infty} \left( \frac{x^{s-1}}{e^x - 1} - \frac{1}{2}x^{s-1}e^{-x} - x^{s-2}e^{-x} \right) \, \mathrm{d}x. \end{align*}

This function is analytic for $\operatorname{Re}(s) > -1$, and $S = S(0)$. Moreover, for $s > 2$, we easily find that

\begin{align*} S(s) &= \Gamma(s)\zeta(s)-\frac{1}{2}\Gamma(s)-\Gamma(s-1) \\ &= \Gamma(s+1)\biggl( \frac{\zeta(s)-\frac{1}{2}-\frac{1}{s-1}}{s} \biggr). \end{align*}

By the principle of analytic continuation, this identity must hold on all of $\operatorname{Re}(s)>-1$. So, letting $s \to 0$ to the above formula yields

$$ S = \lim_{s\to 0}S(s) = 1 + \zeta'(0). $$

Now the desired formula follows from $\zeta'(0) = -\log\sqrt{2\pi}$.

  • $\begingroup$ Maybe a question on its own but how do I find S(s) in terms of gamma and zeta? Somehow I am much more familiar with series expension and summations then integerals. $\endgroup$
    – Gerben
    May 10 '20 at 14:53
  • $\begingroup$ @Gerben, This follows from the formula $$\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,\mathrm{d}x=\sum_{n=1}^{\infty}\int_{0}^{\infty}x^{s-1}e^{-nx}\,\mathrm{d}x=\sum_{n=1}^{\infty}\frac{\Gamma(s)}{n^s}=\Gamma(s)\zeta(s)$$ that holds for $\operatorname{Re}(s)>1$. $\endgroup$ May 10 '20 at 17:51
  • $\begingroup$ @Gerben, But if your goal is to derive the Stirling's formula, then this computation might not be best suited for the purpose, since you will need to derive $\zeta'(0) = -\log\sqrt{2\pi}$ while avoiding a circular argument. $\endgroup$ May 10 '20 at 18:12
  • $\begingroup$ Not the most solid proof like way, but i think i done that ages ago some what intinuative, although i would rephrase the beginning a bit now. math.stackexchange.com/questions/1325169/… $\endgroup$
    – Gerben
    May 10 '20 at 21:30
  • $\begingroup$ And I think i can craft atleast an argument for the integral, I can atleast look into it. $\endgroup$
    – Gerben
    May 10 '20 at 21:33

The series is not convergent, so the formula is wrong. $ (n+\frac 1 2 ) \ln (1+\frac 1 n) \to 1$ as $ n \to \infty$ and this proves that LHS is $\infty$. Also RHS depends on $n$.

  • $\begingroup$ the series does converge i think? You just missed the "- 1" part. It's $(n+\frac{1}[2})\ln(1+\frac{1}{n}-1$ $\endgroup$
    – Gerben
    May 10 '20 at 13:07
  • $\begingroup$ Maybe i should clarify that with big bracelets around the sum. When i compute the imo convergent sum asked, it clearly converge to the value given. Similair for the divergent sum over the negative zeta, but that one diverge once above a certain value, yet for the first few value's it's ofcourse very close to the regularized value given. $\endgroup$
    – Gerben
    May 10 '20 at 13:13

Consider the integral $$\displaystyle \int\limits_{\displaystyle j}^{\displaystyle j+1}\frac{\displaystyle \left \{ x \right \}-\frac{\displaystyle 1}{\displaystyle 2}}{\displaystyle x}dx $$


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