Integral part of Sum

$S = \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{3} } + \dfrac{1}{ \sqrt{4} } + \dots \dots + \dfrac{1}{ \sqrt{9999} } + \dfrac{1}{ \sqrt{10000} }$

Conjecture : $\sqrt{k+1} - \sqrt{k} < \dfrac{1}{( 2 \times \sqrt{k} )} < \sqrt{k} - \sqrt{k-1}$ for positive $k$

Proof : For the left hand side of inequality,

we know that

$0 < \dfrac{1}{4} \Rightarrow k^{2} + k < k^{2} + k + \dfrac{1}{4} \Rightarrow k(k+1) < \bigg ( k + \dfrac{1}{2} \bigg )^{2}$

Since both sides are positive we can square root this to get,

$\sqrt{k(k+1)} < k + \dfrac{1}{2} \Rightarrow \sqrt{k(k+1)} - k < \dfrac{1}{2}$

Dividing both sides by $\sqrt{k}$ we get,

$\sqrt{k+1} - \sqrt{k} < \dfrac{1}{( 2 \times \sqrt{k} )}$

For the right hand side of inequality,

we know that

$0 < \dfrac{1}{4} \Rightarrow k^{2} - k < k^{2} - k + \dfrac{1}{4} \Rightarrow k(k-1) < \bigg ( k - \dfrac{1}{2} \bigg )^{2}$

Since both sides are positive we can square root this to get,

$\sqrt{k(k-1)} < k - \dfrac{1}{2} \Rightarrow k - \sqrt{k(k-1)} > \dfrac{1}{2}$

Dividing both sides by $\sqrt{k}$ we get,

$\dfrac{1}{( 2 \times \sqrt{k} )} < \sqrt{k} - \sqrt{k-1}$

( Motivation : A few days back, I was asked to proof this conjecture in my class. :) )

Now we can write,

$\sqrt{3} - \sqrt{2} < \dfrac{1}{2 \times \sqrt{2}} < \sqrt{2} - \sqrt{1}$

$\sqrt{4} - \sqrt{3} < \dfrac{1}{2 \times \sqrt{3}} < \sqrt{3} - \sqrt{2}$

$\sqrt{5} - \sqrt{4} < \dfrac{1}{2 \times \sqrt{4}} < \sqrt{4} - \sqrt{3}$

$\sqrt{6} - \sqrt{5} < \dfrac{1}{2 \times \sqrt{5}} < \sqrt{5} - \sqrt{4}$

$\dots \dots$

$\dots \dots$

$\dots \dots$

$\sqrt{10000} - \sqrt{9999} < \dfrac{1}{2 \times \sqrt{9999}} < \sqrt{9999} - \sqrt{9998}$

$\sqrt{10001} - \sqrt{10000} < \dfrac{1}{2 \times \sqrt{10000}} < \sqrt{10000} - \sqrt{9999}$

This is like a telescoping series and reduces to,

$\sqrt{10001} - \sqrt{2} < \dfrac{S}{2} < \sqrt{10000} - \sqrt{1}$

$\Rightarrow \sqrt{10001} - \sqrt{2} < \dfrac{S}{2} < 99$

Now, $\sqrt{10000} - \sqrt{2} < \sqrt{10001} - \sqrt{2}$

$\Rightarrow 100 - 1.414213... = 98.58578... < \sqrt{10001} - \sqrt{2}$

This means,

$98.58578..... < \dfrac{S}{2} < 99$

$\Rightarrow 197.171572... < S < 198$

Which means,

$\lfloor S \rfloor = \boxed{197}$

This is mainly a summation of series : $\sum_{i=2}^{10000}\frac{1}{\sqrt{x}}$ This can be written as : $\int\limits_2^{10000}\frac{1}{\sqrt{x}}\mathrm{d}x$

By integrating :

$2\left.\sqrt{x}\right|_2^{10000}$

By evaluating $\left \lfloor{197.17157287525382 }\right \rfloor$ answer is $\boxed{197}$ .

For positive a and b, $\frac { 1 }{ \sqrt { a } } -\frac { 1 }{ \sqrt { b } } =\frac { \sqrt { b } -\sqrt { a } }{ \sqrt { ab } }$ , which is positive if and only if b>a. Hence, $f\left( x \right) =\frac { 1 }{ \sqrt { x } }$ is a monotonically decreasing function for positive x.

$\int{\frac{1}{\sqrt{x}}}=2\sqrt{x}$

$S < \int_{1}^{10000}{\frac{1}{\sqrt{x}}}=2\sqrt{10000}-2\sqrt{1}=200-2=198$ because the right riemann sum of a monotonically decreasing function is an underestimation.

$S > \int_{2}^{10000}{\frac{1}{\sqrt{x}}}=2\sqrt{10000}-2\sqrt{2}>200-2(1.5)=200-3=197$ because the left riemann sum of a monotonically decreasing function is an overestimation.

Therefore, $197 < S < 198$ , so $\left\lfloor S \right\rfloor =\boxed { 197 }$ .

I used this well-known inequality

2sqrt{ $n+1$ } - 2sqrt{ $n$ } < 1/sqrt $n$ < 2sqrt{ $n$ } - 2sqrt{ $n-1$ } 2sqrt{ $m+1$ } - 2sqrt{ $k$ } < sigma[i=k to m] 1/sqrt $n$ < 2sqrt{ $m$ } - 2sqrt{ $k-1$ }

then, we know that $S$ = 1/sqrt $2$ + 1/sqrt $3$ + ... + 1/sqrt $10000$ = sigma[i=2 to 10000], substitute the value $s$ to the inequality we get

2sqrt{ $10001$ } - 2sqrt{ $2$ } < sigma[i=2 to 10000] 1/sqrt $n$ < 2sqrt{ $10000$ } - 2sqrt{ $1$ } --> 2sqrt{ $10001$ } - 2sqrt{ $2$ } < 1/sqrt $2$ + 1/sqrt $3$ + ... + 1/sqrt $10000$ < 2sqrt{ $10000$ } - 2sqrt{ $1$ } --> 197,18157 < S < 198

So the value of floor{ $S$ } is 197

The summation can be written as

$S = \sum\limits_{i=2}^{10000} i^{-\frac{1}{2}}$

Now we can approximate it my representing it as a definite integral which is written as follows:

$\int\limits_{2}^{10000} x^{-\frac{1}{2}} \,dx = 2 \sqrt{x} |_{2}^{10000} \approx 197.17$

So the floor of that value is $\boxed{197}$ .

I honestly feel like I just got lucky with that solution.

Using the Riemann sums for the definite integrals of 1/sqrt(x) from 2 to 10000 & from 2 to 10001 gives us an inequality of S Hence we deduce that the answer is 197

IntegralPartofSum.py

s = 0

for n in range(2,10001):
    s += 1/n**0.5

print int(s)

Returns: 197. Q.E.D.