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Divergent Möbius series

Let \mu denote the Möbius function. Prove that

    \[\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left| \mu (n) \right|}{n} = +\infty\]

Solution

Summing only over primes , where |\mu(p)|=1 ,  we have that

    \[\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left| \mu (n) \right|}{n} \geq \sum_{p \in \mathcal{P}} \frac{1}{p} = +\infty\]

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Application of extended binomial theorem

For the values of a for which the following series makes sense, prove that

    \[\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} = \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )}\]

Solution

We have successively:

    \begin{align*} \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} &= \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\left ( a+n \right )! \left ( a-n \right )!} \\ &=\frac{1}{\left (2a \right )!}\sum_{n=-\infty}^{\infty} \binom{2a}{a+n} \cos nx \\ &=\frac{1}{\left (2a \right )!}\mathfrak{Re} \left ( \sum_{n=-\infty}^{\infty} \binom{2a}{a+n} e^{inx} \right ) \\ &= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\left ( 2a \right )!} \\ &= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )} \end{align*}

due to the extended binomial theorem.

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Double summation

Evaluate the sum

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

Solution

Lemma: It holds that

    \[\sum_{m=1}^{\infty} \frac{1}{\sinh^2 m \pi} = \frac{1}{6} - \frac{1}{2\pi}\]

Proof: Consider the function \displaystyle  f(z)=\frac{\cot \pi z}{\sinh^2 \pi z} and let us it integrate over the following contour \gamma

By the residue theorem it follows that

    \[\oint \limits_{\gamma} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res}_{z_k} f(z)\]

For the residues we have

    \[\begin{matrix} \displaystyle \mathfrak{Res}_{z=n}{\frac{\cot\pi z}{\sinh^2\pi z}} &= & \displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=ni}{\frac{\cot\pi z}{\sinh^2\pi z}} &= &\displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=0}{\frac{\cot\pi z}{\sinh^2\pi z}} & = &\displaystyle -\frac{2}{3\pi} \end{matrix}\]

The integrals along the sides vanish; hence:

    \[-2i\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=2\pi i\left(-\frac{2}{3\pi}+\frac{4}{\pi}\sum_{n=1}^\infty\frac{1}{\sinh^2\pi n}\right)\]

and since

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=\left[ \frac{1}{\pi}\tanh\pi x \right]_{-\infty}^\infty=\frac{2}{\pi}\]

the result follows. \blacksquare

Back to the problem. We have successively:

    \begin{align*} \mathcal{S} &= \frac{1}{2} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2}\sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \prod_{n=1}^{\infty} \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \frac{\sinh \pi j }{\pi j} \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \left ( \frac{1}{j^2} - \frac{\pi^2}{\sinh^2 \pi j} \right ) \\ &= \frac{\pi}{4} \end{align*}

 

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Arithmotheoretic sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^\infty\frac{1}{n}\sum_{d\mid n} \frac{d}{n+d^2}\]

Solution

The sum converges absolutely , so we can switch the order of summation; hence:

    \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \sum_{d\mid n} \frac{d}{n+d^2} &= \sum_{d=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{dk} \frac{d}{dk+d^2} \\ &= \sum_{d=1}^{\infty} \frac{1}{d} \sum_{k=1}^{\infty} \frac{1}{k(d+k)} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^2} \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{d+k} \right) \\ &= \sum_{d=1}^{\infty} \frac{\mathcal{H}_d}{d^2} \end{align*}

The last sum equals 2\zeta(3) and hence

    \[\mathcal{S} = 2\zeta(3)\]

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Convergence of a series

Examine the convergence of a series

    \[\sum_{n=1}^{\infty} \left ( 1 - 2\exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) \right )\]

Solution

From Taylor’s theorem with integral remainder we have that

    \[-\ln 2 = \sum_{k=1}^{n} \frac{(-1)^k}{k} + (-1)^{n+1} \int_0^1 \frac{x^n}{1+x} \, \mathrm{d}x\]

However it is known that \displaystyle \int_{0}^{1} \frac{x^n}{1+x} \, \mathrm{d}x = \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right ). Hence,

    \[\sum_{k=1}^{n} \frac{(-1)^k}{k} = -\ln 2 + \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Exponentiating we get

    \[1- 2 \exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) = \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Thus the series converges.

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