AIEEE 2008 (Offline)Question 1
STATEMENT - 1: For every natural number $n\ge2$, $\frac{1}{\sqrt1}+\frac{1}{\sqrt2}+\ldots+\frac{1}{\sqrt n}>\sqrt n$.
STATEMENT - 2: For every natural number $n\ge2$, $\sqrt{n(n+1)}<n+1$.
Sequence and Series · Miscellaneous
Solution
Formulas used. (1) For $k<n$: $\sqrt k<\sqrt n$, so $\frac1{\sqrt k}>\frac1{\sqrt n}$(2) $n(n+1)<(n+1)^2$Step 1 - Statement 1. Each of the $n$ terms is at least $\dfrac1{\sqrt n}$, and the first $n-1$ are strictly bigger (as $n\ge2$): $$\frac1{\sqrt1}+\cdots+\frac1{\sqrt n}>n\cdot\frac1{\sqrt n}=\sqrt n .$$ Statement 1 is true. Step 2 - Statement 2. $n(n+1)<(n+1)^2$, so $\sqrt{n(n+1)}<n+1$. True. Step 3 - Explanation? Statement 1 is proved by the simple comparison above. Even an induction proof would need the lower bound $\sqrt{n(n+1)}>n$, not the upper bound in Statement 2. So Statement 2 does not explain Statement 1. Answer: (C)