3
$\begingroup$

Does the limit $\displaystyle\lim_{n→∞}\left(\sum_{k=1}^n\dfrac1{\sqrt k}-2\sqrt n\right)$ exist? With integral test I could make this bound: $$-2<\sum_{k=1}^n\frac1{\sqrt k}-2\sqrt n\le-1.$$ How would you tackle this problem?

$\endgroup$
3
$\begingroup$

Assuming you know the generalized harmonic numbers $$\sum_{k=1}^n \dfrac{1}{\sqrt{k}}=H_n^{\left(\frac{1}{2}\right)}$$ Now, using the asymptotics of them $$H_n^{\left(\frac{1}{2}\right)}=2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ making $$\sum_{k=1}^n \dfrac{1}{\sqrt{k}}-2\sqrt{n}=\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ where $\zeta \left(\frac{1}{2}\right)\approx -1.46035$

Let us try for $n=10$; the "exact" calculation would give $-1.30356$ while the approximation would give $-1.30224$.

$\endgroup$
  • $\begingroup$ So, the limit is $\zeta \left( \frac{1}{2} \right)$ ? $\endgroup$ – ARahman Oct 8 '18 at 10:45
  • 1
    $\begingroup$ @ARahman. Yes, it is ! Cheers. $\endgroup$ – Claude Leibovici Oct 8 '18 at 10:53
2
$\begingroup$

Put $$a_n = \sum_{k=1}^n \frac1{\sqrt{k}} -2 \sqrt{n}.$$ We have $$a_{n+1} -a_n = \frac1{\sqrt{n+1}} -2(\sqrt{n+1} -\sqrt{n}) = \frac1{\sqrt{n+1}} -\frac{2}{\sqrt{n} + \sqrt{n+1}} = -\frac{\sqrt{n+1} -\sqrt{n}}{\sqrt{n+1}(\sqrt{n+1} +\sqrt{n})}< 0.$$ Then $a_n$ is decreasing sequence. Moreover $a_n > -2$. Hence there exists the limit $\lim_{n\to \infty} a_n$.

$\endgroup$

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.