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Sinc series

The sinc function is defined as:

    \[\mathrm{sinc} \; x =\left\{\begin{matrix} 1 & , & x=0 \\\\ \dfrac{\sin x}{x} &, & x \neq 0 \end{matrix}\right.\]

Prove that for any couple (\alpha, \beta) of real numbers in (0, 1) the following result holds:

    \[\sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta = \frac{\pi}{\max\{\alpha, \beta \}}\]

Solution

Since

    \[\sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta = 1 + \frac{2}{\alpha \beta} \sum_{n=1}^{\infty} \frac{\sin n \alpha \sin n \beta}{n^2}\]

due to the addition formulas for the sine and cosine functions it is enough to prove the equality

    \[f(\theta) = \sum_{n=1}^{\infty} \frac{\cos n \theta}{n^2} = \frac{\pi^2}{6} - \frac{\theta\left ( 2 \pi - \theta \right )}{4}  \quad \text{forall} \quad \theta \in [0, 2\pi]\]

which is an immediate consequence of the Fourier series by integrating the sawtooth wave function;

    \[\sum_{n=1}^{\infty} \frac{\sin n \theta}{n} = \frac{\pi-\theta}{2}\]

Hence,

    \begin{align*} \sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta &= 1+ \frac{f\left ( \left | \alpha- \beta\right | \right) + f\left ( \alpha+\beta \right )}{\alpha \beta} \\ &=1+\frac{1}{4\alpha \beta} \bigg ( \left ( \alpha-\beta \right )^2 - \left ( \alpha+\beta \right )^2 +\\ &\quad \quad \quad + 2\pi \left ( \alpha-\beta-\left | \alpha+\beta \right | \right ) \bigg ) \\ &= 1+ \frac{1}{4\alpha \beta} \left ( -4\alpha \beta + 4 \pi \min \{ \alpha, \beta \} \right )\\ &= \frac{\pi}{\max\{\alpha, \beta \}} \end{align*}

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Double summation

Let \alpha>3 be a real number. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha}\]

Solution

Since the summands are all positive , we can sum by triangles. Thus,

    \begin{align*} \mathcal{S}&=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha} \\ &=\sum_{n=2}^{\infty} \frac{\sum \limits_{k=1}^{n-1} k}{ n^\alpha} \\ &=\frac{1}{2} \sum_{n=2}^{\infty} \frac{n\left ( n-1 \right )}{n^\alpha} \\ &= \frac{1}{2} \left ( \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-2}} - \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-1}} \right ) \\ &= \frac{\zeta\left ( \alpha-2 \right ) - \zeta\left ( \alpha-1 \right )}{2} \end{align*}

where \zeta is the Riemann zeta function.

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Floor series

Let \left \lfloor \cdot \right \rfloor denote the floor function. Evaluate the series

    \[\mathcal{S} = \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)}\]

Solution

First of all we note that 4k+3 and 4k+2 are never squares. Thus, there exists a positive integer m such that

    \[m^2 \leq \sqrt{4k+1} < \sqrt{4k+2}< \sqrt{4k+3} < \left ( m+1 \right )^2\]

It is easy to see that \sqrt{4x+1} \leq\sqrt{x} + \sqrt{x+1} < \sqrt{4x+3} and thus we conclude that

    \[\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor = \left \lfloor 4k+1 \right \rfloor\]

Now \left \lfloor 4k+1 \right \rfloor is equal to the even number 2n if-f

    \[\left ( 2n \right )^2 \leq 4k+1 < \left ( 2n+1 \right )^2 \Leftrightarrow k \in \left [ n^2, n^2+n \right )\]

Hence, since the series is absolutely convergent we can rearrange the terms and by noting that the finite sums are telescopic , we get that:

    \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)} &= \sum_{n=1}^{\infty} \bigg ( \sum_{k=n^2}^{n^2+n-1} \left ( \frac{1}{k} - \frac{1}{k+1} \right )- \\ &\quad \quad \quad  \quad \quad - \sum_{k=n^2+n}^{\left ( n+1 \right )^2-1}\left ( \frac{1}{k} - \frac{1}{k+1} \right ) \bigg )\\ &=\sum_{n=1}^{\infty} \left ( \frac{1}{n^2} -\frac{1}{n^2+n}-\frac{1}{n^2+n} +\frac{1}{(n+1)^2} \right ) \\ &=\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} - 2 \sum_{n=1}^{\infty} \frac{1}{n^2+n} \\ &=\frac{\pi^2}{6} + \frac{\pi^2}{6}-1 -2 \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\ &=\frac{\pi^2}{3} -1-2 \\ &= \frac{\pi^2}{3}-3 \end{align*}

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Double “identical” series

Compute the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \sum_{k=n+1}^{\infty} \frac{(-1)^{k+1}}{2k-1}\]

Solution

Successively we have:

    \begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \sum_{k=n+1}^{\infty} \frac{(-1)^{k+1}}{2k-1} &= \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \left ( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{2k-1} + \sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \right )\\ &=\sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \left ( \frac{\pi}{4} + \sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \right ) \\ &= -\frac{\pi^2}{16} + \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\sum_{k=1}^{n} \frac{(-1)^k}{2k-1} \\ &= -\frac{\pi^2}{16} + \frac{1}{2} \bigg ( \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\sum_{k=1}^{\infty} \frac{(-1)^k}{2k-1} +\\ & \quad \quad \quad + \sum_{k=1}^{\infty} \frac{1}{\left ( 2k-1 \right )^2} \bigg ) \\ &= \frac{\pi^2}{32} \end{align*}

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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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