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Let $n\in \mathbb N$ and: $$ x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n $$ Show that $\{x_n\}$ is a bounded sequence.

This sequence appears a bit tricky because it involves harmonic series. Below are steps I take.

Lower bound:

$$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \ge \sqrt{\sum_{k=1}^n\left(\frac{k}{k}\right)^2} = \sqrt{\sum_{k=1}^n1}=\sqrt{n}\implies\\ \implies x_n \ge \sqrt n - \sqrt n \ge 0 $$

Lower bound is simple.

Upper bound: To get rid of radical lets use Cauchy-Schwarz (note the below is incorrect as shown in Zvi's answer): $$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \le \sqrt{\left(\sum_{k=1}^n\left(\frac{k+1}{k}\right)\right)^2} =\sum_{k=1}^n\left(\frac{k+1}{k}\right) = n+\sum_{k=1}^n{1\over k} = n + H_n $$

So this doesn't show $x_n$ is bounded above. I've tried another approach:

$$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n} $$ Consider nominator:

$$ \sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=n+\sum_{k=1}^n{2\over k}+\sum_{k=1}^n{1\over k^2}-n = \sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k} $$

For denominator: $$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n = \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n $$

So $x_n$ is: $$ x_n = \frac{\sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}}{ \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n } $$

But i don't see how to proceed from this point. What else could i try? How to show $x_n$ is bounded above?

Please note the precalculus tag.

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  • $\begingroup$ Where does that $\pi^2/6$ come from? Your sequence seems does not involve the infinite sum $\sum_1^\infty 1/k^2 = \pi^2/6$. $\endgroup$
    – xbh
    Nov 2, 2018 at 15:32
  • $\begingroup$ @xbh Right! my bad, the sum involves only $n$ terms, i will update the post $\endgroup$
    – roman
    Nov 2, 2018 at 15:35
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    $\begingroup$ Nominator is at most $\frac{\pi^2}{6}+2\ln(n)+2$. The denominator is at least $\sqrt{n}$. This means we can bound sequence from above by $\frac{\frac{\pi^2}{6}+2\ln(n)+2}{\sqrt{n}}$, converging to $0$ $\endgroup$
    – Jakobian
    Nov 2, 2018 at 15:38

3 Answers 3

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Jakobian gave a very good hint. I am giving somewhat sharper bounds for $x_n$. Note that the OP incorrectly used the Cauchy–Bunyakovsky–Schwarz inequality. Indeed, when applied correctly, we get $$\sqrt{n}\left(x_n+\sqrt{n}\right)=\sqrt{\sum_{k=1}^n1^2}\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}\geq \sum_{k=1}^n1\cdot\left(\frac{k+1}{k}\right)=n+H_n,$$ where $H_n=\sum_{k=1}^n\frac1k$ is the $n$th harmonic number. That is, $$x_n\geq \frac{H_n}{\sqrt{n}}.$$

For the upper bound, the OP got $$x_n=\frac{\sum_{k=1}^n \frac{1}{k^2}+2H_n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}+\sqrt{n}}.$$ We can show that $\sum_{k=1}^n\frac{1}{k^2}$ is bounded above by $2$ via $$\sum_{k=1}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\left(1-\frac{1}{n}\right)<2.$$ However, a sharper upper bound is $\sum_{k=1}^n\frac{1}{k^2}<\zeta(2)=\frac{\pi^2}{6}$. Also, $$\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2\geq \sum_{k=1}^n1^2=n.$$ This proves that $$x_n<\frac{\frac{\pi^2}{6}+2H_n}{\sqrt{n}+\sqrt{n}}=\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$ (You can replace $\frac{\pi^2}{6}$ by $2$ if you want a precalculus solution.)

That is, we have $$\frac{H_n}{\sqrt{n}}\leq x_n<\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$ So, $x_n$ has the asymptotic behavior of $\frac{H_n}{\sqrt{n}}\approx \frac{\ln n}{\sqrt{n}}$. But to show that $x_n$ is bounded, we don't need to know that $H_n\approx \ln n$ (well, to be even more precise, $H_n\approx \gamma+\ln n$, where $\gamma$ is the Euler–Mascheroni constant). Clearly, $$\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right)\leq \frac{\pi^2}{6}\left(\frac{1}{2}\right).$$ Furthermore, $$H_n=\sum_{k=1}^n\frac{1}{k}\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}<\sum_{k=1}^n\frac{2}{\sqrt{k}+\sqrt{k-1}}=2\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}.$$ Hence, $$x_n<\frac{2\sqrt{n}}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2}\right)=2+\frac{\pi^2}{6}\left(\frac{1}{2}\right)<\infty.$$ Again, replace $\frac{\pi^2}{6}$ by $2$ if you don't want any non-precalculus knowledge in the proof. So, you would get $x_n<3$ for all $n$.

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  • $\begingroup$ I've been working through your answer. There is one bit i still don't get. That is $\sqrt n(x_n + \sqrt n) = \dots$. Could you please explain why we need this exact expression and where is it derived from? Also why is $\sum_{k=1}^n {1\over \sqrt k} < \sum_{k=1}^n \frac{2}{\sqrt k + \sqrt{k-1}}$? The rest seems clear to me. Thank you very much for taking your time, your answer is very helpful. I want to accept it once i fully grasp the details $\endgroup$
    – roman
    Nov 5, 2018 at 10:30
  • $\begingroup$ I've understood the second part from my comment, you have multiplied and divided by $2$. $\frac{1}{\sqrt k} = \frac{2}{2\sqrt k} = \frac{2}{\sqrt k + \sqrt k}< \dots$ $\endgroup$
    – roman
    Nov 5, 2018 at 10:44
  • $\begingroup$ The point is I wanted to get the term $\sqrt{\sum_{k=1}^n1^2}\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}$ to use the CBS inequality. But by definition of $x_n$, this expression equals $\sqrt{n}(x_n+\sqrt{n})$. $\endgroup$
    – user593746
    Nov 5, 2018 at 13:37
  • $\begingroup$ Great! Got it now, thank you! $\endgroup$
    – roman
    Nov 5, 2018 at 13:52
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Even if $\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n + 2H_n + H_n^{(2)}$ involves (generalized) harmonic numbers, it is pretty straightforward to notice$^{(*)}$ that $H_n=O(\log n)$ and $H_{n}^{(2)}=O(1)$ as $n\to +\infty$, hence

$$ \sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n\left(1+O\left(\frac{\log n}{n}\right)\right),\tag{1}$$

$$ \sqrt{\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2} = \sqrt{n}\left(1+O\left(\frac{\log n}{n}\right)\right)=\sqrt{n}+O\left(\frac{\log n}{\sqrt{n}}\right)\tag{2}$$ and the wanted limit is zero.

$(*)$ $H_n \leq 1+\int_{1}^{n}\frac{dx}{x}=1+\log(n)$ and $H_{n}^{(2)}\leq 1+\sum_{k=1}^{n}\frac{1}{k(k+1)}\leq 2$.

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You have, using that $a^2-b^2=(a+b)(a-b)$, \begin{align} \sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=\sum_{k=1}^n \left[ \left(1+\frac1k\right)^2-1\right] =\sum_{k=1}^n \frac1k\left(2+\frac1k\right)\leq3\sum_{k=1}^n\frac1k. \end{align} Using the comparison with the integral we get $$ \sum_{k=1}^n\frac1k\leq1+\int_1^n\frac1t\,dt=1+\log n. $$ Then $$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}\leq\frac{3(1+\log n)}{\sqrt n}. $$ Since $\lim_{x\to\infty}\frac{3(1+\log x)}{\sqrt x}=0$, the continous function $f(x)=\frac{3(1+\log x)}{\sqrt x}$ is bounded on $[1,\infty)$. Thus there exists $C>0$ with $$ \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n \leq\frac{3(1+\log n)}{\sqrt n}\leq C. $$

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