show that $$\dfrac{\sqrt{2}}{2}<f(n)=\dfrac{\sqrt{2n+1}-1}{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}}<\dfrac{\sqrt{3}}{2}$$

I know this $$\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}<\dfrac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$$

By the way I can use Stolz lemma find the $$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{\dfrac{1}{\sqrt{n}}}=\dfrac{\sqrt{2}}{2}$$

but this can't prove this inequality,Thank you

  • $\begingroup$ I'm getting a feeling that our target shall remain to simplify the denominator first! This will help us to go further from this point. $\endgroup$ – Kushashwa Ravi Shrimali Jun 25 '14 at 13:04
  • $\begingroup$ If you rationalize each term in the denominator, you get: $$ 1 + \cfrac{\sqrt{2}}{2} + \cfrac{\sqrt{3}}{3} + \dots + \cfrac{\sqrt{n}}{n} $$ $\endgroup$ – Kushashwa Ravi Shrimali Jun 25 '14 at 13:05

The estimate $1/\sqrt{n} < 2(\sqrt{n}-\sqrt{n-1})$ yields

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

and so

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

For an upper bound, we could look at the analogous $1/\sqrt{n} > 2(\sqrt{n+1}-\sqrt{n})$, which yields

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

That would leave

$$\frac{\sqrt{2n+1}-1}{\sqrt{n+1}-1} < \sqrt{3}$$

to be shown, or

$$\sqrt{3}-1 < \sqrt{3(n+1)}-\sqrt{2n+1} = \frac{n+2}{\sqrt{3(n+1)}+\sqrt{2n+1}}.$$

A trivial estimate shows the right hand side is $> \frac{\sqrt{n+2}}{\sqrt{3}+\sqrt{2}}$, which yields the inequality for $n \geqslant 7$.

The cases $1 \leqslant n \leqslant 6$ can be verified by hand.

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  • $\begingroup$ Does estimating like that is considered to be the best way? Just asking, no offense. $\endgroup$ – Kushashwa Ravi Shrimali Jun 25 '14 at 13:20
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    $\begingroup$ The "best" way is always the easiest that gives you the bounds you want/need. So what is the best way depends. For me, at this time, estimating like that was the best way. If one knows a slick inequality that one can apply to the expression, that would probably be considered a better way. $\endgroup$ – Daniel Fischer Jun 25 '14 at 13:24
  • $\begingroup$ That's a positive response Daniel. I appreciate it. Thank You! And Excellent Work. :) $\endgroup$ – Kushashwa Ravi Shrimali Jun 25 '14 at 13:42

Let's start from that Stolz lemma. We see that it converges down to $\frac{\sqrt{2}}{2}$, and we feel that expression under limit in the middle of equality it must be monotonically decreasing. Let's check that.

So, let's $$g(n)=\frac{\sqrt{2n+1} - \sqrt{2n-1}}{\frac{1}{\sqrt{n}}}.$$ Continue it on to real line: $$g(x)=\frac{\sqrt{2x+1} - \sqrt{2x-1}}{\frac{1}{\sqrt{x}}}.$$ Now we calculate $$g'(x)=\frac{-\frac{4 x}{\sqrt{2 x-1}}+\frac{4 x}{\sqrt{2 x+1}}+\frac{1}{\sqrt{2 x-1}}+\frac{1}{\sqrt{2 x+1}}}{2 \sqrt{x}}.$$ Our goal now is to prove that this expression is less that zero $\forall\,x>0.$ It's kinda trivial.

So we have now that $g'(x)<0\ \ \forall\,x>0$, therefore $g(x)$ is decreasing, therefore $g(n)$ is decreasing when $n>1$. Then, I cannot right now strictly prove that the same can we say about $\ f(n)$, but I'm pretty sure you can get some inspiraton in the proof of Stolz lemma (beautiful lemma in my opinion). So let's say that $f(n)$ is monotonically decreasing to $\frac{\sqrt{2}}{2}$. Now we only got to prove that $f(1)$ satisfies your inequality, and then every single $f(n)$ will be jailed below $\frac{\sqrt{3}}{2}$ monotonically converging down to $\frac{\sqrt{2}}{2}$. This particular case is trivial too.

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