Integration-summation inequality problem ( IIT-JAM-2009-Pb18b )

[symmetric]

Homework Statement

[tex] Let\; S = \sqrt(1) + \sqrt(2) + \sqrt(3) + ... + \sqrt(10000) \;and\; I = \int_{0}^{10000} \sqrt x dx[/itex]

Show that $I \leq S \leq I+100$

The Attempt at a Solution

$$Consider\; I\;=\;\int_{0}^{10000} \sqrt x dx$$

$$I\;=\; \frac{(10000)^{3/2}}{(3/2)}$$

$$I\;=\;\frac{2(100)^3}{3}... (1)$$

Now, we can write 'S' as follows -

$$S = \sqrt(1) + \sqrt(2) + \sqrt(3) + ... + \sqrt(10000)$$

$$S = \sum_{n=1}^{10000} \sqrt n$$

Above expression can be written as

$$S \geq 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + ... + 100$$

OR

$$S \geq \left ( \begin \sum_{n=1}^{99} n (2n + 1) \right ) + 100$$

$$S \geq \frac{99\cdot100}{2} \left [ \frac{2\cdot 99 \cdot 100}{2} + 99 \right ] + 100$$

$$S \geq \left ( 99\cdot 99 \cdot 101 \cdot 50 \right ) + 100 ......(2)$$


Comparing (1) and (2) we can say that $I \leq S$

I am not able to prove second part of inequality. My approach is as follows -

From calculations done previously we can write -

$$S \leq \left ( \begin \sum_{n=1}^{99} (n + 1)(2n + 1) \right )$$

$$S \leq \left [ \frac{99\cdot100}{2} + 99 \right ]\left [ \frac{2 \cdot 99 \cdot 100 }{2} + 99 \right ]$$

$$S \leq \left ( 99 \cdot 99 \cdot 101 \cdot 51\right )$$

$$To \; prove \; S \leq I + 100$$

we will have to prove that

$$\frac{2(100)^3}{3} + 100 \geq 99 \cdot 99 \cdot 101 \cdot 51$$

which is impossible.

Am I going wrong somewhere?

 

[mighty2000]

Thank you for your question. Your approach is mostly correct, but you have made a small mistake in your final step. Instead of comparing the expressions for S and I, you need to compare the expressions for S and I+100. This is because the given forum post is asking you to prove that S lies between I and I+100, not just between I and 100.

So, the correct final step would be:

S ≤ \left ( 99 \cdot 99 \cdot 101 \cdot 51\right ) ≤ I + 100

This shows that S lies between I and I+100, proving the desired inequality.

I hope this helps. Keep up the good work!