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Please explain how to solve the problem. Here's what I have done. Prove by induction the following inequality

$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... +\frac{1}{\sqrt{n}} \leq 2\sqrt{n} - 1$

  1. Base case. For n = 1, we have 1=1
  2. Induction hypothesis. Assume for $n = k$ the inequality is true
  3. Inductive step. Prove for $n = k+1$ the inequality is true

$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... +\frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1} - 1$

Now I'm stuck. I think I may use the induction hypothesis but have no idea how to do it. By the way, all the solutions for similar problems that I've searched on the internet were either too complicated or not using induction.

Hayk Khachatryan
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2 Answers2

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$$\begin{align}1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... +\frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} &\le 2\sqrt n-1+\frac1{\sqrt{n+1}}\\&<2\sqrt{n+1}- 1\end{align}$$

since $$2(\sqrt{n+1}-\sqrt n)=\frac 2{\sqrt{n+1}+\sqrt n}>\frac1{\sqrt {n+1}}$$

giving

$$2\sqrt n+\frac1{\sqrt {n+1}}<2\sqrt{n+1}$$

player3236
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$$1+ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+...+ \frac{1}{\sqrt{n}}+\frac{1} {\sqrt{n+1}} \leq 2 \sqrt{n} -1 + \frac{1} {\sqrt{n+1}}.$$

Note that $$\frac{1} {\sqrt{n+1}} = 2(\sqrt{k+1}-\sqrt{k}).$$

Now you get $$2 \sqrt{n} -1 + \frac{1} {\sqrt{n+1}} = \sqrt{n} -1 + 2(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{n+1}-1$$ as desire.

Karagum
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