Floor of a sum

197.

Inequality - 2(√(k+1)-√k)<1√k<2(√k-√(k-1))

(ISI 2005 qsn) I've done this sum earlier in goiit.

yes.but this is a popular exercise in LMVT.

Oh yeah [3]

Inequality can be proved simply from a²>0 for any real a kind-of-argument....

Ok. This has a trick.

Using LMVT on f(x) = \sqrt{x} we get

2(\sqrt{n+1}- \sqrt{n})< \frac{1}{\sqrt{n}}<2(\sqrt{n}- \sqrt{n-1})

So, 2\sum_{n=1}^{10000}{}(\sqrt{n+1}- \sqrt{n})< \sum_{n=1}^{10000}{}\frac{1}{\sqrt{n}}<2\sum_{n=1}^{10000}{}(\sqrt{n}- \sqrt{n-1})

Gives me 198<2(\sqrt{10001}- 1)<\sum_{n=1}^{10000}{}\frac{1}{\sqrt{n}}<200 ,not good enough to find floor of the sum.We need stronger bounds.

To do so we use a stronger inequality \sum_{n=2}^{10000}{}\frac{1}{\sqrt{n}}<2(100-1)

Now adding 1 both sides \sum_{n=1}^{10000}{}\frac{1}{\sqrt{n}}<2(100-1)+1 = 199

Now we have 198<\sum_{n=1}^{10000}{}\frac{1}{\sqrt{n}}<199. Hence \lfloor{\sum_{n=1}^{10000}{}\frac{1}{\sqrt{n}}}\rfloor = \boxed{198}

The desired inequality could be obtained as follows:
\dfrac{1}{\sqrt{n}} = \dfrac{2}{2\sqrt{n}} < \dfrac{2}{\sqrt{n}+\sqrt{n-1}} = 2(\sqrt{n} -\sqrt{n-1})
Similarly
\dfrac{1}{\sqrt{n}} = \dfrac{2}{2\sqrt{n}} >\dfrac{2}{\sqrt{n+1}+\sqrt{n}} = 2(\sqrt{n+1} -\sqrt{n})
And hence
2(\sqrt{n+1} -\sqrt{n}) < \dfrac{1}{\sqrt{n}} < 2(\sqrt{n} -\sqrt{n-1})

@ nasiko, to obtain those tighter bounds I used another inequality, which can be derived from this inequality [3]

I'll try to post it as soon as possible.

BTW are there any such inequalities for reciprocals of the naturals (and not of their roots)?

As far as I remember Anant sir gave 1 qsn long back which involved an inequality in the reciprocals of the squares of naturals....(It had a very dangerous proof though using complex nos. and all that) :D

1) OK.

2)Yes

ln(x+1) - ln(x) <\frac{1}{x}< ln(x)- ln(x+1)

3)I want to see the qsn given by Anant sir long back. :-/

divergent sum