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Basically, I'm trying to prove (by induction) that:

$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}} > 2\:(\sqrt{n+1} − 1)$$

I know to begin, we should use a base case. In this case, I'll use $1$. So we have:

$$1 > 2\:(1+1-1) = 1>-2$$

Which works out.

My problem is the next step. What comes after this? Thanks!

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    $\begingroup$ For $n=1$, the inequation should be $1>2(\sqrt2-1)$ which is indeed true, but not that trivial. Anyway, neither $2(1+1-1)=1$ is valid, and don't know where that $-2$ is coming from... $\endgroup$
    – Berci
    Feb 25, 2013 at 12:03
  • $\begingroup$ The defacing of this user's questions continues... This time, the modifications are absurd. $\endgroup$
    – Did
    May 6, 2013 at 19:35
  • $\begingroup$ Rolled back. Read the rules wrong, my bad! $\endgroup$
    – tekman22
    May 6, 2013 at 20:16

4 Answers 4

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Mean Value Theorem can also be used,

Let $\displaystyle f(x)=\sqrt{x}$

$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$

Using mean value theorem we have:

$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$

$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$....(1)

$\displaystyle \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n}}$

Using the above ineq. in $(1)$ we have,

$\displaystyle \frac{1}{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$

Adding the left part of the inequality we have,$\displaystyle\sum_{k=2}^{n}\frac{1}{2\sqrt{k}}<\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-1$

$\Rightarrow \displaystyle\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2(\sqrt{n}-1)$

$\Rightarrow \displaystyle1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}-2+1=2\sqrt{n}-1$

$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<2\sqrt{n}-1$

Similarly adding the right side of the inequality we have,

$\displaystyle\sum_{k=1}^{n}\frac{1}{2\sqrt{k}}>\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1$

$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}>2(\sqrt{n+1}-1)$

This completes the proof.

$\displaystyle 2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$

This is a much better proof than proving by induction(Ofcourse if someone knows elementary calculus).

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    $\begingroup$ Nice! The mean value theorem (or just the idea of using functions) would never have crossed my mind. $\endgroup$
    – genepeer
    Feb 25, 2013 at 12:03
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Let $f(x) = \frac{1}{\sqrt{x}}$. This is a decreasing function on $\mathbb{R}^+_*$, hence $$ f(k) > \int_k^{k+1} f(x) \, \mathrm{d}x. $$ Summing for $k=1, \dots, n$ give $$ f(1) + f(2) + \dots + f(n) > \int_1^{n+1} f(x) \, \mathrm{d}x,$$ which is exactly the desired inequality.

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Hint(for induction):

$$2(\sqrt{n+1}-1)+\dfrac{1}{\sqrt{n+1}}>2(\sqrt{n+2}-1)\iff \\ 2n+3>2\sqrt{(n+2)(n+1)}\iff \\ (2n+3)^2>4(n+2)(n+1)=(2n+3)^2-1\cdots$$

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We can prove that for $x \geqslant 1$ \begin{align*}\sum_{1\leqslant n\leqslant x}\frac{1}{\sqrt{n}}&\overset{*}=\frac{[x]}{\sqrt{x}}+\frac{1}{2}\int\limits_1^x\frac{[t]}{t^{3/2}}\,dt\\ &=\frac{x+\mathcal{O}(1)}{\sqrt{x}}+\frac{1}{2}\int\limits_1^x\frac{t-\{t\}}{t^{3/2}}\,dt\\ &=\sqrt{x}+\mathcal{O}\left(\frac{1}{\sqrt{x}}\right)+\frac{1}{2}\int\limits_1^x\frac{dt}{\sqrt{t}}-\frac{1}{2}\int\limits_1^{\infty}\frac{\{t\}}{t^{3/2}}\,dt+\frac{1}{2}\int\limits_x^{\infty}\frac{\{t\}}{t^{3/2}}\,dt\\ &=2\sqrt{x}-1-\frac{1}{2}\int\limits_1^{\infty}\frac{\{t\}}{t^{3/2}}\,dt+\mathcal{O}\left(\frac{1}{\sqrt{x}}\right)\\ &=2\sqrt{x}+c+\mathcal{O}\left(\frac{1}{\sqrt{x}}\right),\end{align*} where $\displaystyle c\overset{\text{def}}{=}-1-\frac{1}{2}\int\limits_1^{\infty}\frac{\{t\}}{t^{3/2}}\,dt $ is some absolute real constant.

$*$ - we used here Abel's summation formula for $ a_n=1 $ and $ \phi(n)=\dfrac{1}{\sqrt{n}} $

$[\,\,]$ denotes floor function and $ \{\,\,\} $ fractional part function

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