Prove, using the principle of induction, that for all $n \in \mathbb{N}$, we have have the following inequality: $1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$
An inequality for all natural numbers
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0The inequality, as it currently reads 13:15 Pacific, is false for $n=1,2.$ At 13:18 seems to be correct. – 2012-10-24
3 Answers
Try this:
The inequality hold for $k=1$, assume it holds for $k=n$. Now look at $k=n+1$: $ S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}} $ The last step is the assumption. So LHS is $2\sqrt{n} + \frac{1}{\sqrt{n+1}}$ and RHS is $2 \sqrt{n+1}$ (induction). Clearly $ 2 \sqrt{n+1} - 2 \sqrt{n}=2(\sqrt{n+1} - \sqrt{n})=\frac{2}{\sqrt{n+1}+\sqrt{n}} >\frac{1}{\sqrt{n+1}} $ Therefore, $ S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}<2 \sqrt{n+1} $
HINT
First prove that $2 \sqrt{n} + \dfrac1{\sqrt{n+1}} < 2\sqrt{n+1}$.
To prove this note that $\sqrt{n+1} - \sqrt{n} = \dfrac1{\sqrt{n+1} + \sqrt{n}} > \dfrac1{2\sqrt{n+1}}$
Now couple this with what you have at your induction step.
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0Sorry, i don't understand. Could you prove it? – 2012-10-24
Suppose $1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$ and $1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1}$ (i.e., that the induction hypothesis is false).
Subtracting these, $\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1} - 2\sqrt n = 2(\sqrt {n+1} - \sqrt n)\frac{\sqrt {n+1} +\sqrt n}{\sqrt {n+1} +\sqrt n} = \frac{2}{\sqrt {n+1} +\sqrt n} $ or $\sqrt {n+1} +\sqrt n > 2 \sqrt {n+1}$, which is false.
So the induction hypothesis is true (once a base case is established).