Zeta function reciprocal limit

Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} \frac{\zeta(n+1)}{\zeta(n)}\]

Solution

We note that

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

Hence

    \[1 < \zeta(n) < \frac{1}{1-2^{1-n}}\]

Thus \lim \limits_{n \rightarrow +\infty} \zeta(n) =1. The limit follows to be 1.

Note: It holds that \zeta(n+1) \sim \zeta(n).

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ζ(3) tail in floor function

Let n \in \mathbb{N}. Prove that

    \[\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1)\]

Solution

We note that

    \begin{align*} \sum_{k=n}^{\infty} \frac{1}{k^3} &< \sum_{k=n}^{\infty} \frac{1}{k(k^2-1)} \\ &= \sum_{k=n}^{\infty} \left \frac{1}{2k(k-1)} - \frac{1}{2k(k+1)}\right) \\ &= \frac{1}{2n(n-1)} \end{align*}

as well as

    \begin{align*} \sum_{k=n}^{\infty} \frac{1}{k^3} &> \sum_{k=n}^{\infty} \frac{k}{k^4 + \frac{1}{4}} \\ &= \sum_{k=n}^{\infty} \left( \frac{1}{2k^2 -2k + 1} - \frac{1}{2(k+1)^2 - 2(k+1) + 1}\right) \\ & =\frac{1}{2n^2 - 2n + 1} \\ &= \frac{1}{2n(n-1) + 1} \end{align*}

The result follows.

Note: It also holds that

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

Using the same exact technique we see that

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

as well as

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

The result follows.

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Multiple logarithmic integral

Prove that

    \[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y,z)}{ \ln x + \ln y + \ln z} = - \frac{1}{2}\]

Solution

First of all we observe that the integral \bigintsss_0^1 \frac{\mathrm{d}x}{\ln x} as well as the integral \bigintsss_0^1 \bigintsss_0^1 \frac{\mathrm{d}(x, y)}{\ln x + \ln y} diverge whereas the proposed integral converges which is an interesting fact. Now,

    \begin{align*} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln x + \ln y + \ln z} &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln xyz} \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{u=xyz \; , \; v=y \; , \; w=z}{=\! =\! =\! =\! =\! =\! =\! =\! =\!=\!=\!} \iiint \limits_{\mathbb{D}} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=\int_{0}^{1} \int_{u}^{1} \int_{u/w}^{1} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=-\frac{1}{2} \end{align*}

since \mathbb{D} is determined by the inequalities

    \[\frac{u}{w}\leq v\leq 1 \quad , \quad u\leq w\leq 1\quad ,\quad 0\leq u\leq 1\]

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A beautiful Gamma series

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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On a limit with summation

Let S_k(n) = \sum \limits_{m=1}^{n} m^k \; , \; k \in \mathbb{N}^*. Prove that

    \[\lim_{n \rightarrow +\infty} n \left ( \frac{S_k(n)}{n^{k+1}} - \frac{1}{k+1} \right ) = \frac{1}{2}\]

Solution

Let us begin with the simple observation that

    \begin{align*} \lim_{n \rightarrow +\infty} \frac{1}{n^{k+1}} \sum_{m=1}^{n} m^k &=\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{m=1}^{n} \left ( \frac{m}{n} \right )^k \\ &=\int_{0}^{1} x^k \, \mathrm{d}x \\ &= \frac{1}{k+1} \end{align*}

Now, here comes a handy lemma.

Lemma: Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative. It holds that

    \[\lim_{n\rightarrow +\infty}n \left(\frac{1}{n}\sum_{i=1}^{n}f \left(\frac{i}{n} \right)-\int_{0}^{1}f(x) \, \mathrm{d}x \right)= \frac{f(1)-f(0)}{2}\]

Proof: The derivation of the theorem follows from application of MVT in the interval \left[x , \frac{i}{n} \right].

Hence the limit follows to be \frac{1}{2}.

Note: Applying Euler  – MacLaurin we have

    \[\sum_{m=0}^{n} m^k = \frac{n^{k+1}}{k+1} + \frac{n^k}{2} + \frac{k n^{k-1}}{12} + \mathcal{O}\left(n^{k-3}\right)\]

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