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On the factorial

Let \mu denote the Möbius function and \left \lfloor \cdot \right \rfloor denote the floor function. Prove that:

    \[n! = \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)}\]

Solution

The RHS equals

    \begin{align*} \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)} &= \prod_{k=1}^n \prod_{i|k} \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\mu(i)} \\ &= \prod_{k=1}^n \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\sum_{i|k}\mu(i)} \\ &= n! \end{align*}

since \sum \limits_{i | k} \mu(i) = 0 for k>1.

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On Euler’s totient function series

Let \phi denote Euler’s totient function. Prove that for s>2 it holds that:

    \[\sum_{n=1}^{\infty} \frac{(-1)^n \phi(n)}{n^s} =-\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-3}{2^s-1}\]

where \zeta stands for the Riemann zeta function.

Solution

Well by Euler’s product we have,

    \[\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_{p}\left(1+\frac{\phi(p)}{p^s}+\frac{\phi(p^2)}{p^{2s}}+\frac{\phi(p^3)}{p^{3s}}+\cdots\right)= \prod_{p}\frac{p^s-1}{p^s-p}\]

thus,

(1)   \begin{equation*} \sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_p \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}  \end{equation*}

and

(2)   \begin{equation*} \sum_{\substack{n\geq 1\\n\text{ odd}}}\frac{\phi(n)}{n^s} = \prod_{p>2} \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-2}{2^s-1}  \end{equation*}

Combining we get the result.

Note: It also holds that

    \[\sum_{n=1}^\infty \frac{\phi(2n-1)}{(2n-1)^s} = \frac{(1-2^{1-s})\zeta(s-1)}{1-2^{-s} \zeta(s)}\]

 

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Inverse zeta(3) limit

Evaluate the limit

    \[\ell= \lim_{T \rightarrow +\infty} \frac{1}{2T} \int \limits_{-T}^{T} \frac{\zeta(\frac{3}{2}+it)}{\zeta(\frac{3}{2}-it)} \, {\rm d}t\]

Solution

We are proving that the limit is \frac{1}{\zeta(3)}. Indeed , one has:

    \[\frac{\zeta(3/2+it)}{\zeta(3/2-it)} = \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) (d^2/n)^{it}\]

If x \neq 1 then

    \[\lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T x^{it}dt = \lim_{T \to +\infty} \frac{1}{2T} \frac{x^{iT}-x^{-iT}}{i\ln x} = 0\]

whereas if x=1 then the above limit is 1. Thus:

\begin{aligned} \lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T \frac{\zeta(3/2+it)}{\zeta(3/2-it)} dt &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) \lim_{T \to +\infty} \frac{1}{2T}\int_{-T}^T (d^2/n)^{it}dt \\ &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) 1_{n = d^2} \\ &= \frac{1}{\zeta(3)} \end{aligned}

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On a strange Möbius series

Let \mu denote the Möbius function. Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}\]

where \mathfrak{Re}(s)>1.

Solution

Since

    \[(-1)^{\mu(n)}= 1 - 2\mu^2(n)\]

we deduce that

\displaystyle \sum_{n= 1}^{\infty} \frac{(-1)^{\mu(n)}}{n^s} = \zeta(s)-2\sum_{n\geq 1}\frac{\mu^2(n)}{n^s} = \zeta(s)-\frac{2\,\zeta(s)}{\zeta(2s)}

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An analytic logarithmic sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=-\infty}^{\infty} \frac{\log \left | n + \frac{1}{4} \right |}{n + \frac{1}{4}}\]

(Seraphim Tsipelis)

Solution [pprime]

We have successively:

\begin{aligned} \sum_{n=-\infty}^{\infty} \frac{\log \left | n + \frac{1}{4} \right |}{n + \frac{1}{4}} &= 4 \log \frac{1}{4} + \sum_{n=1}^{\infty} \left [ \frac{\log \left ( n + \frac{1}{4} \right )}{n + \frac{1}{4}} - \frac{\log \left ( n - \frac{1}{4} \right )}{n - \frac{1}{4}} \right ] \\ &= - 8 \log 2 + 8 \log 2 \sum_{n=1}^{\infty} \left [ \frac{1}{4n-1} - \frac{1}{4n+1} \right ] +\\ &\quad \quad +4 \sum_{n=1}^{\infty} \left [ \frac{\log(4n+1)}{4n+1} - \frac{\log(4n-1)}{4n-1} \right ] \\ &=-8 \log 2 + 8 \log 2 \left ( 1 - \frac{\pi}{4} \right ) + \\ &\quad \quad \quad +4\sum_{n=1}^{\infty} \frac{(-1)^n \log (2n+1)}{2n+1} \\ &= - 2 \pi \log 2 + 4 \sum_{n=1}^{\infty} \frac{\log n}{n} \sin \frac{n \pi}{2} \end{aligned}

We are invoking Kummer’s formula for the evaluation of the last sum. Evaluating the Fourier series that appear for z=\frac{1}{4} we get that

    \[\mathcal{S}= - \pi \cdot \left( {4 \cdot \log 2 - 4 \cdot \log \Gamma \left( {\dfrac{1}{4}} \right) + 3 \cdot \log \pi + \gamma } \right)\]

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