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Convergence of a series

May 12, 2020 / 1 Comment on 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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Factorial series

April 26, 2020 / 1 Comment on Factorial series

Evaluate the series

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{1}{(3n)!}\]

Solution

Let \delta_{n, k} denote Kronecker’s delta and \zeta_m = e^{2\pi m i /3}. We have successively,

    \begin{align*} \sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n}}{k!} \\ &=\frac{1}{2\pi i } \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\; \oint \limits_{\left | z \right |=R>1} z^{3n-k+1} \, \mathrm{d} z\\ &= \frac{1}{2\pi i }\oint \limits_{ \left | z \right |=R>1} \frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{3n}} \sum_{k=0}^{\infty} \frac{z^k}{k!} \, \mathrm{d} z\\ &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=R>1} \frac{e^z}{z \left ( 1-\frac{1}{z^3} \right )}\, \mathrm{d} z\\ &= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=R>1} \frac{e^z z^2}{z^3-1} \, \mathrm{d}z \\ &=\sum_{m=-1}^{1}\frac{\zeta_m^2 e^{\zeta_m}}{3\zeta_m^2} \end{align*}

It follows that

    \[\mathcal{S} =\frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}\]

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Otto Dunkel Memomorial

April 17, 2020 / Leave a comment

Prove that

    \[\sum_{n=1}^{\infty}\frac{1}{n}\int_{2n\pi}^{\infty}\frac{\sin z}{z}\,\mathrm{d}z=\pi-\frac{\pi \ln 2 \pi}{2}\]

Solution

We have successively:

    \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \int_{2\pi n}^{\infty} \frac{\sin z}{z} \,\mathrm{d}z &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{\sin nt}{t+2\pi} \, \mahrm{d}t \\ &=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \int_{0}^{2\pi} \frac{\sin nt}{t + 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \sum_{n=1}^{\infty} \frac{1}{n} \frac{\sin nt}{t+ 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \frac{\pi -t}{2\left (t + 2 m\pi \right )} \, \mathrm{d}t\\ &= \pi \sum_{n=1}^{\infty} \left [ \left ( 1 + \frac{1}{2} \right ) \log \left ( 1 + \frac{1}{n} \right ) -1 \right ] \\ &=\pi \log \left [ \lim_{N \rightarrow +\infty} e^{-N} \prod_{n=1}^{N} \left ( \frac{n+1}{n} \right )^{n+1/2} \right ] \\ &= \pi \log \left ( \lim_{N \rightarrow +\infty} \frac{\sqrt{N+1}\left ( N+1 \right )^N e^{-N}}{N!} \right ) \\ &= \pi \log \left ( \frac{e}{\sqrt{2\pi}} \right ) \\ &= \pi - \frac{ \pi \log 2\pi}{2} \end{align*}

since for x \in (0, 2\pi) it holds that

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

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A squared trigamma series

April 17, 2020 / Leave a comment

Prove that

    \[\sum_{n=1}^{\infty} \left( \psi^{(1)} (n) \right)^2 = 3 \zeta(3)\]

Solution

Since \displaystyle \psi^{(1)}(n) = \sum_{k=n}^{\infty} \frac{1}{k^2} we have successively:

    \begin{align*} \sum_{n=1}^\infty \left( \psi^{(1)}(n) \right)^2 &=\sum_{n=1}^\infty\sum_{j=n}^\infty\frac1{j^2}\sum_{k=n}^\infty\frac1{k^2} \\ &=\sum_{n=1}^\infty\left(\sum_{j=n}^\infty\frac1{j^4}+2\sum_{j=n}^\infty\sum_{m=1}^\infty\frac1{j^2}\frac1{(j+m)^2}\right) \\ &=\sum_{j=1}^\infty\sum_{n=1}^j\frac1{j^4}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^j\frac1{j^2}\frac1{(j+m)^2} \\ &=\sum_{j=1}^\infty\frac1{j^3}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\frac1{j(j+m)^2} \\ &=\zeta(3)+2\sum_{n=1}^\infty\frac{\mathcal{H}_{n-1}}{n^2} \\ &=\zeta(3)-2\zeta(3)+2\sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} \\ &=\zeta(3)-2\zeta(3)+4\zeta(3) \\ &=3\zeta(3) \end{align*}

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Pierre Mounir series

April 14, 2020 / Leave a comment

Let \kappa>0. Prove that

    \[\sum_{n=1}^{\infty} \frac{n \mod \kappa}{n^2+n} = \ln \kappa\]

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