10
$\begingroup$

Show that $$ L=\lim_{s\rightarrow\infty}\left(\int_0^s\frac{ds'}{\sqrt{s'}}-\sum_{s'=1}^s\frac{1}{\sqrt{s'}}\right) = 1.460\ldots $$

My attempts: To begin, rewriting the limit of the form $$ L=\lim_{\epsilon\rightarrow0}\left(\int_0^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}ds'-\sum_{s'=1}^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}\right) $$ where $$ \int_0^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}ds' = \int_0^{\infty}\frac{e^{-\epsilon s^2}}{s}d(s^2) = \sqrt{\frac{\pi}{\epsilon}} $$ and $$ \sum_{s'=1}^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}} = \sum_{s'=1}^{\infty}{e^{-\epsilon s'}}\int_0^{\infty}e^{-\sqrt{s'}t}dt = \int_0^{\infty}\left(\sum_{s'=1}^{\infty}{e^{-\epsilon s'-\sqrt{s'}t}}\right)dt $$ Second, the identity $$ \sum_{s=1}^{\infty}\frac{(1-\epsilon)^s}{\sqrt{s}}=\sqrt{\frac{\pi}{\epsilon}}(1+O(\epsilon)); $$ may be of some help.

Thirdly, the limit is somehow $-\zeta(1/2)=1.46035\cdots$ where $\zeta(s)$ is the Riemann zeta function.

$\endgroup$
  • $\begingroup$ $\displaystyle{\sum_{s=1}^{\infty} \frac{{(1-\epsilon)}^{s}}{\sqrt{s}}} = \sum_{s=1}^{\infty} \frac{1-s\epsilon}{\sqrt{s}} = \zeta{(1/2)}-\sum_{s=1}^{\infty} \frac{s\epsilon}{\sqrt{s}}$ $\endgroup$ – Oussama Boussif Sep 11 '15 at 9:14
8
$\begingroup$

Let us represent the finite sum $\sum_{k=1}^n\frac{1}{\sqrt{k}}$ as \begin{align*}\sum_{k=1}^n\frac{1}{\sqrt{k}}&=\sum_{k=1}^n\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{e^{-kx}dx}{\sqrt x}=\frac{1}{\sqrt \pi}\int_0^{\infty} e^{-x}\frac{1-e^{-(n+1)x}}{1-e^{-x}}\frac{dx}{\sqrt x}=\\ &=\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x+\frac{e^{-x}}x\right)dx=\\&= \frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx+\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{e^{-x}-e^{-(n+2)x}}{x\sqrt x}dx=\\ &=\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx+2\sqrt{n+2}-2. \end{align*} Since $\sqrt{n+2}-\sqrt n\to 0$ as $n\to \infty$, the limit we are looking for is given by $$L=2-\frac{1}{\sqrt \pi}\int_0^{\infty} \left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)\frac{dx}{\sqrt x}.$$ It is not difficult to relate this expression to zeta value $\zeta\left(\frac12\right)$. Indeed, the integral $$I(s)=\frac{1}{\sqrt \pi}\int_0^{\infty} x^{s-1}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx$$ converges and defines an analytic function of $s$ in the region $\Re s>0$. Furthermore for $\Re s>1$ we can break it into two separate pieces, which leads to evaluation \begin{align*}I(s)=\color{blue}{\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{x^{s-1} dx}{e^x-1}}-&\color{red}{\frac{1}{\sqrt \pi}\int_0^{\infty}x^{s-2}e^{-x}dx}=\frac{\color{blue}{\Gamma(s)\zeta(s)}-\color{red}{\Gamma(s-1)}}{\sqrt\pi}=\\ &=\frac{\Gamma(s)}{\sqrt\pi}\left[\zeta(s)-\frac{1}{s-1}\right]. \end{align*} This finally gives $L=2-I\left(\frac12\right)=-\zeta\left(\frac12\right)$.

$\endgroup$
1
$\begingroup$

I suppose that you want to show that the sequence $u_n=2\sqrt{n}-\sum_{k=1}^n \frac{1}{\sqrt{k}}$ is convergent. We compute $v_n=u_{n+1}-u_n$. I have found $$v_n=\frac{1}{(\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})^2}$$ As the series $\sum v_n$ is convergent, the sequence $u_n$ is convergent, and the limit $L$ is $L=1+\sum_{k\geq 1} v_k$.

$\endgroup$
  • 1
    $\begingroup$ The question is how to compute that limit $\endgroup$ – Oussama Boussif Sep 11 '15 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.