Question

Let $S = \frac{1}{\sqrt{1}+\sqrt{3}} + \frac{1}{\sqrt{5}+\sqrt{7}} + \frac{1}{\sqrt{9}+\sqrt{11}} + ... + \frac{1}{\sqrt{9997}+\sqrt{9999}}$, then

• A. S < 24
• B. S > 24
• C. S > 18
• D. S < 18

Answer 1

Answer: (B)

S > 24

Answer 2

The given sum is $S = \frac{1}{\sqrt{1}+\sqrt{3}} + \frac{1}{\sqrt{5}+\sqrt{7}} + \frac{1}{\sqrt{9}+\sqrt{11}} + ... + \frac{1}{\sqrt{9997}+\sqrt{9999}}$. The general term of the series is $T_n = \frac{1}{\sqrt{4n-3}+\sqrt{4n-1}}$. There are 2500 terms in the sum.

First, rationalize the general term $T_n$: $T_n = \frac{1}{\sqrt{4n-3}+\sqrt{4n-1}} = \frac{\sqrt{4n-1}-\sqrt{4n-3}}{(\sqrt{4n-1}+\sqrt{4n-3})(\sqrt{4n-1}-\sqrt{4n-3})} = \frac{\sqrt{4n-1}-\sqrt{4n-3}}{(4n-1)-(4n-3)} = \frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}$

So the sum $S$ can be written as: $S = \sum_{n=1}^{2500} \frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}$

Let $f(x) = \sqrt{x}$. By the Mean Value Theorem, for $a < b$, there exists a $c \in (a,b)$ such that $\frac{f(b)-f(a)}{b-a} = f'(c)$. Here, let $a = 4n-3$ and $b = 4n-1$. Then $b-a = 2$. So, $\frac{\sqrt{4n-1}-\sqrt{4n-3}}{2} = f'(c_n)$ for some $c_n \in (4n-3, 4n-1)$. Since $f'(x) = \frac{1}{2\sqrt{x}}$, we have $T_n = \frac{1}{2\sqrt{c_n}}$. Since $4n-3 < c_n < 4n-1$, we can establish bounds for $T_n$: $\frac{1}{2\sqrt{4n-1}} < T_n < \frac{1}{2\sqrt{4n-3}}$.

Summing these inequalities from $n=1$ to $n=2500$: $\sum_{n=1}^{2500} \frac{1}{2\sqrt{4n-1}} < S < \sum_{n=1}^{2500} \frac{1}{2\sqrt{4n-3}}$

Using integral approximation, we find that $24.57075 < S < 25.24625$.

From this range, we can conclude that $S > 24$.