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Floor series

May 14, 2019 / Leave a comment

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

May 12, 2019 / Leave a comment

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

April 17, 2019 / Leave a comment

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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A hypergeometric series

January 25, 2019 / Leave a comment

Let \alpha, \beta \in \mathbb{R} such that 0<\alpha<\beta. Evaluate the series:

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left (\beta+1 \right )\left ( \beta+2 \right )\cdots\left ( \beta+n \right )}\]

Solution

Lemma 1: For the \mathrm{B}, \Gamma functions , it holds that:

    \[\mathrm{B}(x, y) = \int_{0}^{1} t^{x-1} \left ( 1-t \right )^{y-1}\, \mathrm{d}t = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}\]

Lemma 2: It holds that:

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}= \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\]

Proof: Simple calculations using Lemma (1) reveal the identity.

Lemma 3: Using Lemma 2 it holds that

    \[\mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right ) = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

and as a consequence

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )} = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

 

Then, successively we have that:

\begin{aligned} \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left ( \beta+1 \right )\left ( \beta+2 \right )\cdots \left ( \beta+n \right )} &= \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right )_n}{\left ( \beta+1 \right )_n} \\ &=\sum_{n=0}^{\infty} \frac{\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \alpha \right )}}{\frac{\Gamma\left ( \beta+n+2 \right )}{\Gamma\left ( \beta \right )}} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right )} \cdot \sum_{n=0}^{\infty} \frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)}\int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \sum_{n=0}^{\infty} t^n \, \mathrm{d}t \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \cdot \frac{\mathrm{d}t}{1-t} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha-1} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \mathrm{B} \left ( \alpha+1, \beta-\alpha \right ) \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \frac{\Gamma\left ( \alpha+1 \right ) \Gamma\left ( \beta-\alpha \right )}{\Gamma\left ( \beta+1 \right )} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \left( \beta -\alpha \right) \cdot \Gamma \left(\beta-\alpha \right)} \cdot \frac{\alpha \Gamma \left (\alpha \right ) \Gamma\left ( \beta - \alpha \right )}{\beta \Gamma \left ( \beta \right )} \\ &= \frac{\alpha}{\beta} \cdot \frac{1}{\beta-\alpha} \end{aligned}

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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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