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Series of zeta sum

October 20, 2018 / Leave a comment

Let \zeta denote the Riemann zeta function. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( n - \sum_{k=2}^{n} \zeta(k) \right)\]

Solution

Let a \in \mathbb{R} \mid \left| a \right|<2. We are proving the more general result.

    \[\sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) = a \left ( \frac{\psi^{(0)}\left ( 2-a \right ) + \gamma}{1-a} + 1 \right )\]

where \psi^{(0)} denotes the digamma function.

First of all, we note that:

    \begin{align*} \sum_{k=2}^{n} \zeta(k) &= \sum_{k=2}^{n} \sum_{m=1}^{\infty} \frac{1}{m^k} \\ &= \sum_{m=1}^{\infty} \sum_{k=2}^{n} \frac{1}{m^k}\\ &=\sum_{m=1}^{\infty} \sum_{k=0}^{n-2} \frac{1}{m^{k+2}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} \\ &=\sum_{m=1}^{1} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} + \sum_{m=2}^{\infty} \frac{1}{m^2}\sum_{k=0}^{n-2} \frac{1}{m^k} \\ &= n-1 + \sum_{m=2}^{\infty} \frac{1}{m^2} \left ( \frac{1-\frac{1}{m^{n-1}}}{1-\frac{1}{m}} \right ) \\ &= n-1 +\sum_{m=2}^{\infty} \frac{m^{n-1}-1}{\left ( m-1 \right ) m^n} \\ & = n -1 + \sum_{m=1}^{\infty} \frac{1}{m(m+1)} - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \\ &= n - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \end{align*}

Thus,

    \begin{align*} \sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) &= \sum_{m=2}^{\infty} \frac{1}{m-1}\sum_{n=2}^{\infty} \left ( \frac{a}{m} \right )^n \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1) m \left ( m-a \right )} \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{m-a} \left ( \frac{1}{m-1} - \frac{1}{m} \right ) \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1)(m-a)} -a^2 \sum_{m=2}^{\infty} \frac{1}{m(m-a)} \\ &=a^2 \sum_{m=1}^{\infty} \frac{1}{m(m+1-a)} - a^2 \sum_{m=1}^{\infty} \frac{1}{m(m-a)} + \\ & \quad \quad \quad \quad \quad + \frac{a^2}{1-a} \\ &=\frac{a^2}{1-a} \left ( \psi^{(0)} (2-a) + \gamma \right ) +\\ & \quad \quad \quad \quad \quad + a \left ( \psi^{(0)} (1-a) + \gamma \right )+\frac{a^2}{1-a}\\ &= a \left ( \frac{\psi^{(0)}(2-a) + \gamma}{1-a} + 1 \right ) \end{align*}

due to the reflection formula \displaystyle \psi^{(0)} (z+1) = \psi^{(0)}(z) + \frac{1}{z}.

Side note: If a=1 then the sum equals \zeta(2) - 1.

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Harmonic sum with reciprocal central binomial coefficient

July 27, 2018 / Leave a comment

Let \mathcal{H}_n denote the n – th harmonic number. Prove that

    \[\sum_{n=1}^{\infty} \frac{2^n \mathcal{H}_n}{\binom{2n}{n}} = \pi - \frac{\pi \log 2}{2} + \mathcal{G}\]

where \mathcal{G} denotes the Catalan’s constant.

Solution

We begin with a lemma:

Lemma: Let \mathrm{Li}_2 denote the dilogarithm function. It holds that

    \[\mathrm{Li}_2 \left ( \frac{1+i}{2} \right ) - \mathrm{Li}_2 \left ( \frac{1-i}{2} \right ) = -\frac{i}{4} \left ( \pi \log 2 - 8 \mathcal{G} \right )\]

Proof: It is well known that

(1)   \begin{equation*} \mathrm{Li}_2(z) + \mathrm{Li}_2 \left ( \frac{z}{z-1} \right ) = - \frac{1}{2}\ln^2 \left ( 1-z \right ) \quad , \quad z \notin [1, +\infty) \end{equation*}

Setting z=\frac{1+i}{2} we have that:

    \begin{align*} \mathrm{Li}_2 \left ( \frac{1+i}{2} \right ) + \mathrm{Li}_2 \left ( -i \right ) &= -\frac{1}{2} \ln^2 \left ( 1 - \frac{1+i}{2} \right ) \\ &=- \frac{1}{2} \ln^2 \left ( \frac{1-i}{2} \right ) \\ &= -\frac{1}{2} \left ( -\frac{\pi^2}{16} + \frac{\log^2 2}{4} + \frac{i \pi \log 2}{4} \right )\\ &=\frac{\pi^2}{32} - \frac{\log^2 2}{8} - \frac{i \pi \log 2}{8} \end{align*}

Setting z=\frac{1-i}{2} we have that:

    \begin{align*} \mathrm{Li}_2 \left ( \frac{1-i}{2} \right ) + \mathrm{Li}_2 \left ( i \right ) &= -\frac{1}{2} \ln^2 \left ( 1 - \frac{1-i}{2} \right ) \\ &=- \frac{1}{2} \ln^2 \left ( \frac{1+i}{2} \right ) \\ &= -\frac{1}{2} \left ( -\frac{\pi^2}{16} + \frac{\log^2 2}{4} - \frac{i \pi \log 2}{4} \right )\\ &=\frac{\pi^2}{32} - \frac{\log^2 2}{8} + \frac{i \pi \log 2}{8} \end{align*}

However,

    \begin{align*} \operatorname{Li}_2(iz) &= -\int_0^{z} \frac{\log (1-ix)}{x}\,dx \\ &= -\int_0^{z} \frac{\log \left((1+x^2)^{1/2}e^{-i\arctan x}\right)}{x}\,dx \\ &= -\frac{1}{2}\int_0^{z} \frac{\log \left(1+x^2\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\ &= -\frac{1}{4}\int_0^{-z^2} \frac{\log \left(1-x\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\ &= \frac{1}{4}\operatorname{Li}_2(-z^2) + i\operatorname{Ti}_2(z) \end{align*}

where \mathrm{Ti}_2 is the inverse tangent function. It now follows that

    \[\mathrm{Li}_2(i) = -\frac{\pi^2}{48} + i \mathcal{G} \quad , \quad \mathrm{Li}_2(-i) = - \frac{\pi^2}{48} -i \mathcal{G}\]

in view of the well known series \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} = \mathcal{G}.

Substracting the above relations we get the proof of the lemma.

Theorem: Let t=\frac{z}{z-1} \; , \; |t| \leq 1. It holds that

    \[\sum_{n=0}^{\infty} \frac{\mathcal{H}_n 4^n}{\binom{2n}{n}} z^n = \left ( 1+t \right ) \bigg [ \left (2 + \frac{1}{2} \log \frac{1+t}{4} \right ) \sqrt{t} \arctan \sqrt{t} -\]

    \[\hspace{3em} - \frac{\sqrt{-t}}{2} \left ( \mathrm{Li}_2 \left ( \frac{1+\sqrt{-t}}{2} \right ) - \mathrm{Li}_2 \left ( \frac{1-\sqrt{-t}}{2} \right ) \right ) \bigg ]\]

Setting z=\frac{1}{2} we have that t=1. The result now follows immediately using the lemma above as well as the fact that i=\sqrt{-1}.

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On Vacca’s formula

June 2, 2018 / Leave a comment

Let \gamma denote the Euler Mascheroni constant. Prove that

    \[\gamma = \sum\limits_{n=1}^{\infty}(-1)^n\frac{\lfloor{\log_2n}\rfloor}{n}\]

where \lfloor \cdot \rfloor denotes the floor function.

Solution

We have successfully

    \begin{align*} \sum_{n=1}^{2^k-1} (-1)^{n} \frac{\left \lfloor \log_2 n \right \rfloor}{n} &= \sum_{n=1}^{2^k-1} \frac{(-1)^n}{n} \left \lfloor \frac{\log n}{\log 2} \right \rfloor \\ &=\sum_{r=0}^{k-1} \sum_{n=2^r}^{2^{r+1}-1} r\frac{(-1)^n}{n} \\ &= \sum_{r=1}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n} \\ &= \sum_{r=1}^{k-1} r \left( \sum_{n=2^r}^{2^{r+1}-1} \frac{-1}{n} + \sum_{n=2^{r-1}}^{2^r-1} \frac{1}{n}\right) \\ &=1+\sum_{r=1}^{k-2}(r+1-r) \sum_{n=2^r}^{2^{r+1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n}\\ &= \sum_{n=1}^{2^{k-1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n} \\ &\sim \log 2^{k-1} + \gamma + \mathcal{O}(1) - (k-1) \log 2 - \mathcal{O}(1)\\ &= \gamma \end{align*}

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A beautiful Gamma series

May 24, 2018 / Leave a comment

Let \Gamma denote the Gamma function. Prove that

    \[\sum_{n=0}^{\infty} \frac{\Gamma \left ( n + \frac{3}{2} \right )}{(2n+1)(2n+3) (n+1)!} = -\frac{\pi \sqrt{\pi}}{4} + \sqrt{\pi}\]

Solution

The \arcsin Taylor series is

    \[\arcsin x = \sum_{n=0}^{\infty}{\frac{\left( 2n \right) !}{4^n\left( n! \right) ^2\left( 2n+1 \right)}x^{2n+1}}\]

Hence,

\begin{aligned} \sum\limits_{n=0}^{\infty }{\frac{\Gamma \left( n+\frac{3}{2} \right)}{\left( 2n+1 \right)\left( 2n+3 \right)\left( n+1 \right)!}}&=\sum\limits_{n=1}^{\infty }{\frac{\Gamma \left( n+\frac{1}{2} \right)}{\left( 2n-1 \right)\left( 2n+1 \right)n!}}\\ &=\Gamma \left( \frac{1}{2} \right)+\frac{\sqrt{\pi }}{2}\sum\limits_{n=0}^{\infty }{\frac{\left( 2n \right)!}{{{4}^{n}}{{\left( n! \right)}^{2}}}\cdot \frac{2n+1-\left( 2n-1 \right)}{\left( 2n-1 \right)\left( 2n+1 \right)}\\ &=\sqrt{\pi }+\frac{\sqrt{\pi }}{2}\sum\limits_{n=0}^{+\infty }{\frac{\left( 2n \right)!}{{{4}^{n}}{{\left( n! \right)}^{2}}}\left( \frac{1}{2n-1}-\frac{1}{2n+1} \right)}\\ &=\sqrt{\pi }-\frac{\sqrt{\pi }}{2}\arcsin 1 \\ &=\sqrt{\pi }\left( 1-\frac{\pi }{4} \right)} \end{aligned}

and the result follows.

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Series with number of bits ‘1’

April 14, 2018 / 1 Comment on Series with number of bits ‘1’

Let {\rm pop} count the number of bits ‘1’ in the binary representation of n. Calculate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \frac{{\rm pop}(n)}{n(n+1)}\]

Solution

For k \in \mathbb{N}, consider the function:

    \[\theta_k(n) = \left\{\begin{matrix} 1 & , & \text{if k-th bit of n is set} \\ 0 & , & \text{otherwise} \end{matrix}\right.\]

We have:

    \[\sum_{n=1}^{\infty} \frac{\operatorname{pop}(n)}{n(n+1)} = \sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac{\theta_k(n)}{n(n+1)} = \sum_{k=0}^{\infty}\sum_{n=1}^{\infty} \frac{\theta_k(n)}{n(n+1)}\]

because the summands in the double sum are all non-negative numbers and allow us to perform the double sum in any order we want.

Notice \theta_k(n) = 1 iff (2\ell+ 1)2^k \leq n < (2\ell+2)2^k for some integer \ell \in \mathbb{N}. Thus:

    \begin{align*} \sum_{k=0}^{\infty}\sum_{\ell=0}^{\infty} \sum_{n=(2\ell+1)2^k}^{(2\ell+2)2^k-1} \frac{1}{n(n+1)} &= \sum_{k=0}^{\infty}\sum_{\ell=0}^{\infty} \left(\frac{1}{(2\ell+1)2^k} - \frac{1}{(2\ell+2)2^k}\right)\\ &= \left(\sum_{k=0}^{\infty}2^{-k}\right)\sum_{\ell=0}^{\infty} \left(\frac{1}{2\ell+1} - \frac{1}{2\ell+2}\right) \\ &= \frac{1}{1-2^{-1}}\log 2 \\ &= 2\log 2 \end{align*}

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